OP Malhotra Class 11 Maths Solutions Chapter 16 The Straight Line Chapter Test

Question 1. The coordinates of the points A, B, C are (0, 4), (2, 5) and (3, 3) respectively. Prove that AB = BC and the angle ABC is a right angle.
Answer: The coordinates of points are \(A(0, 4)\), \(B(2, 5)\), and \(C(3, 3)\).
First, we find the distance between A and B using the distance formula:
\(AB = \sqrt{(2-0)^2+(5-4)^2}\)
\(AB = \sqrt{2^2+1^2}\)
\(AB = \sqrt{4+1}\)
\(AB = \sqrt{5}\) units.
Next, we find the distance between B and C:
\(BC = \sqrt{(3-2)^2+(3-5)^2}\)
\(BC = \sqrt{1^2+(-2)^2}\)
\(BC = \sqrt{1+4}\)
\(BC = \sqrt{5}\) units.
Since \(AB = \sqrt{5}\) and \(BC = \sqrt{5}\), we have proven that \(AB = BC\).
Now, let's find the distance between A and C:
\(CA = \sqrt{(3-0)^2+(3-4)^2}\)
\(CA = \sqrt{3^2+(-1)^2}\)
\(CA = \sqrt{9+1}\)
\(CA = \sqrt{10}\) units.
To check if angle ABC is a right angle, we use the Pythagorean theorem: \(AB^2 + BC^2 = AC^2\).
\(AB^2 = (\sqrt{5})^2 = 5\)
\(BC^2 = (\sqrt{5})^2 = 5\)
\(AC^2 = (\sqrt{10})^2 = 10\)
So, \(AB^2 + BC^2 = 5 + 5 = 10\).
Since \(AB^2 + BC^2 = AC^2\) (i.e., \(10 = 10\)), the triangle \( \triangle ABC \) is a right-angled triangle, and the right angle is at point B. This shows that the sides fit the Pythagorean relationship perfectly.
In simple words: We calculated the lengths of the sides of the triangle. Sides AB and BC are both the same length, \(\sqrt{5}\). When we square these lengths and add them, we get \(5 + 5 = 10\). This is the same as the square of the third side AC, which is also 10. Because \(AB^2 + BC^2 = AC^2\), it means the angle at B is a right angle (90 degrees).

🎯 Exam Tip: Remember to clearly state the distance formula and the Pythagorean theorem when proving properties of triangles based on coordinates. Show each calculation step for full marks.

Question 2. Find the coordinates of the point which divides the line joining (5, – 2) and (9, 6) in the ratio 3 : 1.
Answer: Let the two given points be \(A(5, -2)\) and \(B(9, 6)\). We need to find the coordinates of a point P that divides the line segment AB in the ratio \(m:n = 3:1\). We will use the section formula for this.

The section formula for a point \(P(x, y)\) dividing a line segment joining \((x_1, y_1)\) and \((x_2, y_2)\) in the ratio \(m:n\) is:
\(P(x, y) = \left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)\)
Here, \((x_1, y_1) = (5, -2)\), \((x_2, y_2) = (9, 6)\), \(m = 3\), and \(n = 1\).
Substitute these values into the formula:
\(P(x, y) = \left(\frac{3 \times 9 + 1 \times 5}{3+1}, \frac{3 \times 6 + 1 \times (-2)}{3+1}\right)\)
\(P(x, y) = \left(\frac{27 + 5}{4}, \frac{18 - 2}{4}\right)\)
\(P(x, y) = \left(\frac{32}{4}, \frac{16}{4}\right)\)
\(P(x, y) = (8, 4)\)
So, the coordinates of point P are \((8, 4)\). This method directly calculates the exact location of the point.
In simple words: To find the point that splits a line in a certain ratio, we use a special formula. We multiply the ratio numbers by the coordinates of the *opposite* points, then add these results, and finally divide by the sum of the ratio numbers. For this problem, point P is at \((8, 4)\).

🎯 Exam Tip: Be careful with the order of multiplication in the section formula: \(m\) is multiplied by \(x_2\) and \(y_2\), and \(n\) is multiplied by \(x_1\) and \(y_1\). A common mistake is to swap them.

Question 3. The vertices of a quad. PMQS are P(0, 0), M(3, 2), Q(7, 7) and S(4, 5). Show that PMQS is a parallelogram.
Answer: The given vertices of the quadrilateral are \(P(0, 0)\), \(M(3, 2)\), \(Q(7, 7)\), and \(S(4, 5)\). To show that PMQS is a parallelogram, we can either prove that its opposite sides are equal in length, or that its diagonals bisect each other (i.e., they share the same midpoint). We will use both methods for a complete proof.

**Method 1: Proving opposite sides are equal**
Calculate the length of each side using the distance formula \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).
Length of PM:
\(PM = \sqrt{(3-0)^2 + (2-0)^2} = \sqrt{3^2 + 2^2} = \sqrt{9+4} = \sqrt{13}\) units.
Length of MQ:
\(MQ = \sqrt{(7-3)^2 + (7-2)^2} = \sqrt{4^2 + 5^2} = \sqrt{16+25} = \sqrt{41}\) units.
Length of QS:
\(QS = \sqrt{(4-7)^2 + (5-7)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9+4} = \sqrt{13}\) units.
Length of SP:
\(SP = \sqrt{(0-4)^2 + (0-5)^2} = \sqrt{(-4)^2 + (-5)^2} = \sqrt{16+25} = \sqrt{41}\) units.
From the calculations, we see that \(PM = QS = \sqrt{13}\) and \(MQ = SP = \sqrt{41}\).
Since both pairs of opposite sides are equal in length, the quadrilateral PMQS is a parallelogram.
**Method 2: Proving diagonals bisect each other**
A quadrilateral is a parallelogram if its diagonals bisect each other. This means they should have the same midpoint. The diagonals are PQ and MS.
Midpoint of diagonal PQ: \(P(0,0)\), \(Q(7,7)\)
Midpoint \( = \left(\frac{0+7}{2}, \frac{0+7}{2}\right) = \left(\frac{7}{2}, \frac{7}{2}\right)\).
Midpoint of diagonal MS: \(M(3,2)\), \(S(4,5)\)
Midpoint \( = \left(\frac{3+4}{2}, \frac{2+5}{2}\right) = \left(\frac{7}{2}, \frac{7}{2}\right)\).
Since the midpoints of both diagonals are the same, the diagonals bisect each other.
Therefore, PMQS is a parallelogram. A parallelogram has the property that its diagonals cross exactly in the middle.
We can also verify using slopes (optional for showing it's a parallelogram, but good for understanding).
Slope of PM \( = \frac{2-0}{3-0} = \frac{2}{3}\)
Slope of QS \( = \frac{5-7}{4-7} = \frac{-2}{-3} = \frac{2}{3}\)
Since slopes are equal, \(PM \parallel QS\).
Slope of MQ \( = \frac{7-2}{7-3} = \frac{5}{4}\)
Slope of SP \( = \frac{5-0}{4-0} = \frac{5}{4}\)
Since slopes are equal, \(MQ \parallel SP\).
Because both pairs of opposite sides are parallel, PMQS is a parallelogram.
In simple words: We checked the lengths of the opposite sides of the shape PMQS. We found that the top and bottom sides are equal, and the left and right sides are also equal. This means it's a parallelogram. We also checked that the middle point of its two main diagonal lines is the same for both. This also confirms it's a parallelogram because diagonals always cut each other in half in a parallelogram.

🎯 Exam Tip: You only need one method (equal opposite sides, or bisecting diagonals, or parallel opposite sides) to prove a quadrilateral is a parallelogram. Choose the one that seems easiest given the coordinates. Showing more than one method demonstrates a deeper understanding.

Question 4. P, Q and R are three collinear points. P and Q are (3, 4) and (7, 7) respectively, and PR = 10 units. Find the coordinates of R.
Answer: Let the coordinates of point R be \((h, k)\).
Since points P, Q, and R are collinear, they lie on the same straight line. This means the area of the triangle formed by these three points is zero.
Area of \(\triangle PQR = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = 0\)
Here, \(P(3, 4)\), \(Q(7, 7)\), \(R(h, k)\).
So, \(\frac{1}{2} |3(7-k) + 7(k-4) + h(4-7)| = 0\)
\(|21 - 3k + 7k - 28 + h(-3)| = 0\)
\(|-7 + 4k - 3h| = 0\)
\(-3h + 4k - 7 = 0\)
\(3h - 4k + 7 = 0\) ... (1)
We are also given that the distance \(PR = 10\) units.
Using the distance formula for P\((3, 4)\) and R\((h, k)\):
\(PR = \sqrt{(h-3)^2 + (k-4)^2}\)
\(PR^2 = (h-3)^2 + (k-4)^2\)
\(10^2 = (h-3)^2 + (k-4)^2\)
\(100 = (h-3)^2 + (k-4)^2\) ... (2)
From equation (1), we can express \(h\) in terms of \(k\):
\(3h = 4k - 7\)
\(h = \frac{4k-7}{3}\)
Substitute this expression for \(h\) into equation (2):
\(100 = \left(\frac{4k-7}{3} - 3\right)^2 + (k-4)^2\)
\(100 = \left(\frac{4k-7-9}{3}\right)^2 + (k-4)^2\)
\(100 = \left(\frac{4k-16}{3}\right)^2 + (k-4)^2\)
\(100 = \frac{(4(k-4))^2}{9} + (k-4)^2\)
\(100 = \frac{16(k-4)^2}{9} + (k-4)^2\)
Factor out \((k-4)^2\):
\(100 = (k-4)^2 \left(\frac{16}{9} + 1\right)\)
\(100 = (k-4)^2 \left(\frac{16+9}{9}\right)\)
\(100 = (k-4)^2 \left(\frac{25}{9}\right)\)
Now, solve for \((k-4)^2\):
\((k-4)^2 = \frac{100 \times 9}{25}\)
\((k-4)^2 = 4 \times 9\)
\((k-4)^2 = 36\)
Take the square root of both sides:
\(k-4 = \pm \sqrt{36}\)
\(k-4 = \pm 6\)
This gives two possible values for \(k\):
Case 1: \(k-4 = 6 \implies k = 4+6 \implies k = 10\)
Case 2: \(k-4 = -6 \implies k = 4-6 \implies k = -2\)
Now, substitute these \(k\) values back into the expression for \(h\): \(h = \frac{4k-7}{3}\)
For \(k=10\):
\(h = \frac{4(10)-7}{3} = \frac{40-7}{3} = \frac{33}{3} = 11\)
So, one possible coordinate for R is \((11, 10)\). This means R can be on one side of P.
For \(k=-2\):
\(h = \frac{4(-2)-7}{3} = \frac{-8-7}{3} = \frac{-15}{3} = -5\)
So, another possible coordinate for R is \((-5, -2)\). This means R can also be on the other side of P.
Therefore, the coordinates of point R are \((11, 10)\) or \((-5, -2)\).
In simple words: We know that points P, Q, and R are on a straight line, and the distance from P to R is 10. We used the "area of a triangle is zero" rule to get one equation for the coordinates of R. Then, we used the distance formula for PR to get another equation. By solving these two equations together, we found two possible locations for R: \((11, 10)\) or \((-5, -2)\).

🎯 Exam Tip: When dealing with collinear points and distances, remember that solving quadratic equations might give two possible solutions. Always substitute both values back into the original equations to ensure they are valid.

Question 5. The coordinates of the vertices of a triangle are (4, – 3), (- 5, 2) and (x, y). If the centroid of the triangle is at the origin, show that x = y = 1.
Answer: Let the vertices of the triangle be \(A(4, -3)\), \(B(-5, 2)\), and \(C(x, y)\).
The centroid of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by the formula:
Centroid \(G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
Substitute the given vertex coordinates into this formula:
Centroid \(G = \left(\frac{4 + (-5) + x}{3}, \frac{-3 + 2 + y}{3}\right)\)
\(G = \left(\frac{4 - 5 + x}{3}, \frac{-1 + y}{3}\right)\)
\(G = \left(\frac{x-1}{3}, \frac{y-1}{3}\right)\)
We are given that the centroid of the triangle is at the origin, which is \((0, 0)\).
So, we can set the coordinates of G equal to \((0, 0)\):
\(\frac{x-1}{3} = 0\)
\(\implies x-1 = 0\)
\(\implies x = 1\)
And,
\(\frac{y-1}{3} = 0\)
\(\implies y-1 = 0\)
\(\implies y = 1\)
Therefore, we have shown that \(x = 1\) and \(y = 1\), which means \(x = y = 1\). The centroid is the balancing point of the triangle.
In simple words: We have a triangle with three corners. We know two corners, and the third one is \((x, y)\). We are also told that the center point of the triangle (called the centroid) is exactly at \((0, 0)\). By using the formula for the centroid, we add up all the x-coordinates and divide by 3, and do the same for the y-coordinates. When we set these sums equal to 0, we find that \(x\) must be 1 and \(y\) must be 1.

🎯 Exam Tip: Remember the centroid formula clearly. When the centroid is at the origin, it simplifies to summing the coordinates to zero. Make sure to apply it separately for x and y components.

Question 6. Find the equation of the line passing through the point (- 4, – 5) and perpendicular to the line joining the points (1, 2) and (5, 6).
Answer: First, let's find the slope of the line joining the points \((1, 2)\) and \((5, 6)\).
The slope \(m\) of a line passing through \((x_1, y_1)\) and \((x_2, y_2)\) is given by \(m = \frac{y_2 - y_1}{x_2 - x_1}\).
Slope of the given line \(= \frac{6-2}{5-1} = \frac{4}{4} = 1\).
Let \(m_1\) be the slope of this line, so \(m_1 = 1\).
The required line is perpendicular to this given line. If two lines are perpendicular, the product of their slopes is -1 (i.e., \(m_1 \times m_2 = -1\)).
So, the slope of the required line (\(m_2\)) is \(m_2 = -\frac{1}{m_1} = -\frac{1}{1} = -1\).
Now we have the slope of the required line (\(m = -1\)) and a point it passes through (\((-4, -5)\)). We can use the point-slope form of a linear equation: \(y - y_1 = m(x - x_1)\).
Substitute the values:
\(y - (-5) = -1(x - (-4))\)
\(y + 5 = -1(x + 4)\)
\(y + 5 = -x - 4\)
Move all terms to one side to get the standard form:
\(x + y + 5 + 4 = 0\)
\(x + y + 9 = 0\)
This is the equation of the required line. It's important to understand how slopes are related in perpendicular lines.
In simple words: We need to find a line that goes through a specific point \((-4, -5)\) and crosses another line at a perfect right angle. First, we found how steep the second line is (its slope). Then, because our line is perpendicular, its steepness is the negative inverse of the first line's steepness. With this new steepness and the point it passes through, we wrote the equation for our line, which is \(x + y + 9 = 0\).

🎯 Exam Tip: Always remember that the slopes of perpendicular lines multiply to -1. If one slope is \(m\), the perpendicular slope is \(-1/m\). Use the point-slope form for the final equation.

Question 7. Find the equation of the straight line which passes through the point (3, 4) and has intercepts on the axes such that their sum is 14.
Answer: Let the equation of the line in intercept form be \(\frac{x}{a} + \frac{y}{b} = 1\) ... (1), where \(a\) is the x-intercept and \(b\) is the y-intercept.
We are given that the sum of the intercepts is 14:
\(a + b = 14\) ... (2)
From equation (2), we can express \(a\) in terms of \(b\): \(a = 14 - b\).
The line passes through the point \((3, 4)\). So, we can substitute \(x=3\) and \(y=4\) into equation (1):
\(\frac{3}{a} + \frac{4}{b} = 1\)
To combine these fractions, find a common denominator:
\(\frac{3b + 4a}{ab} = 1\)
\(3b + 4a = ab\) ... (3)
Now, substitute \(a = 14 - b\) from equation (2) into equation (3):
\(3b + 4(14 - b) = (14 - b)b\)
\(3b + 56 - 4b = 14b - b^2\)
\(56 - b = 14b - b^2\)
Rearrange the terms to form a quadratic equation:
\(b^2 - b - 14b + 56 = 0\)
\(b^2 - 15b + 56 = 0\)
Now, factor the quadratic equation to find the values of \(b\). We need two numbers that multiply to 56 and add to -15 (which are -7 and -8):
\((b - 7)(b - 8) = 0\)
This gives two possible values for \(b\):
\(b = 7\) or \(b = 8\)
Now find the corresponding values for \(a\) using \(a = 14 - b\):
Case 1: If \(b = 7\), then \(a = 14 - 7 = 7\).
Case 2: If \(b = 8\), then \(a = 14 - 8 = 6\).
Therefore, there are two possible equations for the straight line. This happens because the conditions can be met by two different lines.
For Case 1: \(a = 7\), \(b = 7\)
\(\frac{x}{7} + \frac{y}{7} = 1\)
Multiply by 7 to clear the denominators:
\(x + y = 7\)
For Case 2: \(a = 6\), \(b = 8\)
\(\frac{x}{6} + \frac{y}{8} = 1\)
To clear the denominators, multiply by the least common multiple of 6 and 8, which is 24:
\(24 \left(\frac{x}{6}\right) + 24 \left(\frac{y}{8}\right) = 24 \times 1\)
\(4x + 3y = 24\)
So, the two possible equations of the straight lines are \(x + y = 7\) and \(4x + 3y = 24\).
In simple words: We are looking for a line that crosses the x and y axes, and the sum of where it crosses (the intercepts) is 14. Also, the line goes through the point \((3, 4)\). We use the line equation form that has intercepts. We found two possible lines that fit all the rules: \(x + y = 7\) and \(4x + 3y = 24\).

🎯 Exam Tip: When a problem involves intercepts, the intercept form of the line equation \(\frac{x}{a} + \frac{y}{b} = 1\) is often the most efficient starting point. Be prepared for quadratic equations that might yield multiple solutions for intercepts.

Question 8. Reduce the equation of the straight line \(3x + 4y + 14 = 0\) to normal form and find the perpendicular distance of the line from the origin.
Answer: The given equation of the line is \(3x + 4y + 14 = 0\).
To convert this to the normal form \(x \cos \alpha + y \sin \alpha = p\), where \(p\) is the perpendicular distance from the origin and \(p > 0\), we first move the constant term to the right side and ensure it's positive:
\(3x + 4y = -14\)
Multiply by -1 to make the right side positive:
\(-3x - 4y = 14\)
Now, divide the entire equation by \(\pm \sqrt{A^2 + B^2}\), where \(A=-3\) and \(B=-4\). We choose the sign such that \(p\) is positive.
\(\sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)
So, divide the equation \(-3x - 4y = 14\) by 5:
\(\frac{-3x}{5} - \frac{4y}{5} = \frac{14}{5}\)
This is the normal form: \(\left(-\frac{3}{5}\right)x + \left(-\frac{4}{5}\right)y = \frac{14}{5}\).
Comparing this with \(x \cos \alpha + y \sin \alpha = p\), we get:
\(\cos \alpha = -\frac{3}{5}\)
\(\sin \alpha = -\frac{4}{5}\)
\(p = \frac{14}{5}\)
The perpendicular distance of the line from the origin is \(p = \frac{14}{5}\) units.
To find the angle \(\alpha\), we can use \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\):
\(\tan \alpha = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3}\)
Since both \(\cos \alpha\) and \(\sin \alpha\) are negative, \(\alpha\) lies in the third quadrant.
\(\alpha = \pi + \tan^{-1}\left(\frac{4}{3}\right)\). This angle specifies the direction of the normal vector from the origin to the line.
In simple words: We changed the given line equation into a special form called the "normal form." This form helps us find the shortest distance from the origin (the point \((0,0)\)) to the line. After doing the calculations, we found that this distance is \(\frac{14}{5}\) units. The normal form also tells us the angle of this shortest line from the origin.

🎯 Exam Tip: When converting to normal form, ensure the constant term \(p\) is always positive. If it's negative, multiply the entire equation by -1 before dividing by \(\sqrt{A^2+B^2}\). The sign of \(A\) and \(B\) (coefficients of \(x\) and \(y\)) determines the quadrant of \(\alpha\).

Question 9. Find the equation of the straight line which passes through the point of intersection of the straight lines \(x + y = 8\) and \(3x – 2y + 1 = 0\) and is parallel to the straight line joining the points (3, 4) and (5, 6).
Answer: Let the given lines be:
\(L_1: x + y - 8 = 0\) ... (1)
\(L_2: 3x - 2y + 1 = 0\) ... (2)
The equation of any line passing through the intersection of \(L_1\) and \(L_2\) is given by \(L_1 + kL_2 = 0\):
\((x + y - 8) + k(3x - 2y + 1) = 0\)
Group the terms by \(x\) and \(y\):
\(x(1 + 3k) + y(1 - 2k) + (-8 + k) = 0\) ... (3)
The slope of this line (3) is given by \(m = -\frac{\text{coefficient of } x}{\text{coefficient of } y}\):
Slope of line (3) \( = -\frac{(1 + 3k)}{(1 - 2k)} = \frac{1 + 3k}{2k - 1}\).
Now, we need to find the slope of the line joining the points \((3, 4)\) and \((5, 6)\).
Slope of this line \( = \frac{6 - 4}{5 - 3} = \frac{2}{2} = 1\).
Since the required line (3) is parallel to the line joining \((3, 4)\) and \((5, 6)\), their slopes must be equal.
So, set the slopes equal:
\(\frac{1 + 3k}{2k - 1} = 1\)
\(1 + 3k = 2k - 1\)
\(3k - 2k = -1 - 1\)
\(k = -2\)
Now, substitute the value of \(k = -2\) back into equation (3):
\(x(1 + 3(-2)) + y(1 - 2(-2)) + (-8 + (-2)) = 0\)
\(x(1 - 6) + y(1 + 4) + (-8 - 2) = 0\)
\(x(-5) + y(5) - 10 = 0\)
\(-5x + 5y - 10 = 0\)
Divide the entire equation by -5 to simplify:
\(x - y + 2 = 0\)
This is the required equation of the straight line. The parameter \(k\) helps us find the specific line in a family of lines.
In simple words: We needed to find a line that passes through where two other lines cross and is also parallel to a fourth line. First, we wrote a general equation for any line passing through the first two lines' intersection. Then, we found the "steepness" (slope) of the fourth line. Since our new line is parallel to it, they must have the same steepness. We used this to find a special number called \(k\), and when we put \(k\) back into our general equation, we got the final answer: \(x - y + 2 = 0\).

🎯 Exam Tip: For lines passing through the intersection of two lines, use the \(L_1 + kL_2 = 0\) form. Remember that parallel lines have equal slopes. Always simplify the final equation if possible.

Question 10. Find the equation of the straight line which passes through the point of intersection of the straight lines \(3x – 4y + 1 = 0\) and \(5x + y − 1 = 0\) and cuts off equal intercepts from the axes.
Answer: Let the given lines be:
\(L_1: 3x - 4y + 1 = 0\) ... (1)
\(L_2: 5x + y - 1 = 0\) ... (2)
The equation of any line passing through the intersection of \(L_1\) and \(L_2\) is given by \(L_1 + kL_2 = 0\):
\((3x - 4y + 1) + k(5x + y - 1) = 0\)
Group the terms by \(x\) and \(y\):
\(x(3 + 5k) + y(-4 + k) + (1 - k) = 0\)
Rearrange into the form \(Ax + By = C\):
\((3 + 5k)x + (k - 4)y = k - 1\) ... (3)
Now, we need the intercepts of this line on the axes to be equal.
To find the x-intercept, set \(y=0\):
\((3 + 5k)x = k - 1 \implies x = \frac{k-1}{3+5k}\)
To find the y-intercept, set \(x=0\):
\((k - 4)y = k - 1 \implies y = \frac{k-1}{k-4}\)
Since the intercepts are equal, we set them equal to each other:
\(\frac{k-1}{3+5k} = \frac{k-1}{k-4}\)
There are two possibilities for this equality:
1. \(k-1 = 0 \implies k = 1\).
If \(k=1\), substitute into equation (3):
\((3 + 5(1))x + (1 - 4)y = 1 - 1\)
\(8x - 3y = 0\)
This line passes through the origin. Its intercepts are \(0\), which are equal. So, \(8x - 3y = 0\) is one possible equation.
2. If \(k-1 \neq 0\), then we can divide both sides by \((k-1)\):
\(\frac{1}{3+5k} = \frac{1}{k-4}\)
This means the denominators must be equal:
\(3 + 5k = k - 4\)
\(5k - k = -4 - 3\)
\(4k = -7\)
\(k = -\frac{7}{4}\)
Now, substitute \(k = -\frac{7}{4}\) back into equation (3):
\(\left(3 + 5\left(-\frac{7}{4}\right)\right)x + \left(-\frac{7}{4} - 4\right)y = -\frac{7}{4} - 1\)
\(\left(\frac{12 - 35}{4}\right)x + \left(\frac{-7 - 16}{4}\right)y = \frac{-7 - 4}{4}\)
\(\left(-\frac{23}{4}\right)x + \left(-\frac{23}{4}\right)y = -\frac{11}{4}\)
Multiply the entire equation by \(-4\):
\(23x + 23y = 11\)
This is the second required equation. When intercepts are equal, the line often has a simpler form like \(x+y=c\).
In simple words: We want to find a line that goes through the meeting point of two other lines, and it must cut the x and y axes at the same distance from the origin. We first found a general equation for any line passing through the first two lines' intersection. Then, we calculated its x-intercept and y-intercept and set them equal. This gave us two possible values for a special number \(k\). Using these \(k\) values, we found two lines that fit the rules: \(8x - 3y = 0\) and \(23x + 23y = 11\).

🎯 Exam Tip: When dealing with equal intercepts, remember that if the intercept is zero, the line passes through the origin. Also, don't forget to consider the case where the numerator \((k-1)\) is zero, as this can be a valid solution for equal intercepts.

Question 11. Find the locus of a point such that the line segments having end points (2, 0) and (- 2, 0) subtend a right angle at that point.
Answer: Let the given end points of the line segment be \(A(2, 0)\) and \(B(-2, 0)\).
Let \(P(h, k)\) be any point on the locus.
The problem states that the line segments PA and PB subtend a right angle at P. This means that the angle \(\angle APB\) is \(90^\circ\).
According to geometry, if an angle in a semicircle is a right angle, then the hypotenuse is the diameter of the circle. This means point P lies on a circle whose diameter is the line segment AB.

Alternatively, we can use the Pythagorean theorem for \(\triangle APB\): \(PA^2 + PB^2 = AB^2\).
Distance \(PA^2\): \((h-2)^2 + (k-0)^2 = (h-2)^2 + k^2\)
Distance \(PB^2\): \((h-(-2))^2 + (k-0)^2 = (h+2)^2 + k^2\)
Distance \(AB^2\): \((-2-2)^2 + (0-0)^2 = (-4)^2 + 0^2 = 16\)
Substitute these into the Pythagorean theorem:
\((h-2)^2 + k^2 + (h+2)^2 + k^2 = 16\)
Expand the squared terms:
\((h^2 - 4h + 4) + k^2 + (h^2 + 4h + 4) + k^2 = 16\)
Combine like terms:
\(2h^2 + 2k^2 + 8 = 16\)
Move the constant to the right side:
\(2h^2 + 2k^2 = 16 - 8\)
\(2h^2 + 2k^2 = 8\)
Divide the entire equation by 2:
\(h^2 + k^2 = 4\)
To express the locus, we replace \(h\) with \(x\) and \(k\) with \(y\):
\(x^2 + y^2 = 4\)
This is the equation of a circle centered at the origin with a radius of \(\sqrt{4} = 2\). This makes sense because the midpoint of AB is \((0,0)\) and the radius is half the length of AB, which is \(\frac{1}{2} \times 4 = 2\).
In simple words: We are looking for all the points P such that if you draw lines from P to A and P to B, the angle at P is a right angle. We used the distance formula to find the squared lengths of PA, PB, and AB. Then, using the rule that \(PA^2 + PB^2 = AB^2\) for a right-angled triangle, we got the equation \(h^2 + k^2 = 4\). This means the points P form a circle centered at the origin with a radius of 2.

🎯 Exam Tip: When a problem asks for the locus of a point subtending a right angle at a segment, it implies the point lies on a circle whose diameter is that segment. The Pythagorean theorem is a key tool here, but understanding the geometric property can help verify your answer.

Question 12. Find the coordinates of the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (4, 3).
Answer: Let the vertices of the triangle be \(A(1, 2)\), \(B(2, 3)\), and \(C(4, 3)\).
The orthocenter is the point where all three altitudes of a triangle intersect. An altitude is a line segment from a vertex perpendicular to the opposite side.

**1. Find the equation of altitude AL (from A to BC):**
Slope of BC \( = \frac{3-3}{4-2} = \frac{0}{2} = 0\).
Since the slope of BC is 0, BC is a horizontal line. Therefore, the altitude AL must be a vertical line.
A vertical line passing through \(A(1, 2)\) has the equation \(x = 1\) ... (1).
**2. Find the equation of altitude BM (from B to AC):**
Slope of AC \( = \frac{3-2}{4-1} = \frac{1}{3}\).
The altitude BM is perpendicular to AC, so its slope is the negative reciprocal of the slope of AC.
Slope of BM \( = -\frac{1}{\frac{1}{3}} = -3\).
Using the point-slope form \(y - y_1 = m(x - x_1)\) for point \(B(2, 3)\) and slope \(-3\):
\(y - 3 = -3(x - 2)\)
\(y - 3 = -3x + 6\)
\(3x + y - 9 = 0\) ... (2).
**3. Find the equation of altitude CN (from C to AB):**
Slope of AB \( = \frac{3-2}{2-1} = \frac{1}{1} = 1\).
The altitude CN is perpendicular to AB, so its slope is the negative reciprocal of the slope of AB.
Slope of CN \( = -\frac{1}{1} = -1\).
Using the point-slope form \(y - y_1 = m(x - x_1)\) for point \(C(4, 3)\) and slope \(-1\):
\(y - 3 = -1(x - 4)\)
\(y - 3 = -x + 4\)
\(x + y - 7 = 0\) ... (3).
The orthocenter is the intersection point of any two of these altitudes. Let's find the intersection of altitude AL (equation 1) and altitude BM (equation 2).
From (1), we have \(x = 1\).
Substitute \(x = 1\) into (2):
\(3(1) + y - 9 = 0\)
\(3 + y - 9 = 0\)
\(y - 6 = 0\)
\(y = 6\)
So, the point of intersection of AL and BM is \((1, 6)\).
To confirm, let's check if this point also lies on altitude CN (equation 3):
\(x + y - 7 = 0\)
\(1 + 6 - 7 = 0\)
\(7 - 7 = 0\)
\(0 = 0\)
Since \((1, 6)\) satisfies all three altitude equations, it is indeed the orthocenter. It is interesting to note that the orthocenter can be outside the triangle for obtuse triangles.
In simple words: The orthocenter is where all three "altitude" lines of a triangle meet. An altitude is a line from a corner that goes straight down to the opposite side at a right angle. We found the equations for these three lines. Then, we solved for where two of them cross. That crossing point, \((1, 6)\), is the orthocenter of the triangle.

🎯 Exam Tip: To find the orthocenter, you only need to find the equations of two altitudes and then solve them simultaneously to find their intersection point. Verifying with the third altitude is a good way to check your work.

Question 13. Find the equation of the bisector of the acute angle between the lines \(3x – 4y + 7 = 0\) and \(- 12x – 5y + 2 = 0\).
Answer: Let the two given lines be:
\(L_1: 3x - 4y + 7 = 0\)
\(L_2: -12x - 5y + 2 = 0\)
First, ensure the constant terms are positive. \(L_1\) has \(+7\), which is positive. \(L_2\) has \(+2\), which is positive.
The equations of the bisectors of the angles between two lines \(A_1x + B_1y + C_1 = 0\) and \(A_2x + B_2y + C_2 = 0\) are given by:
\(\frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}}\)
For \(L_1: 3x - 4y + 7 = 0\), \(\sqrt{A_1^2 + B_1^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\).
For \(L_2: -12x - 5y + 2 = 0\), \(\sqrt{A_2^2 + B_2^2} = \sqrt{(-12)^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13\).
So, the equations of the angle bisectors are:
\(\frac{3x - 4y + 7}{5} = \pm \frac{-12x - 5y + 2}{13}\)
We need to determine which sign (\(+\) or \(-\)) corresponds to the acute angle bisector. We use the condition \(A_1A_2 + B_1B_2\).
\(A_1 = 3\), \(B_1 = -4\)
\(A_2 = -12\), \(B_2 = -5\)
\(A_1A_2 + B_1B_2 = (3)(-12) + (-4)(-5) = -36 + 20 = -16\).
Since \(A_1A_2 + B_1B_2\) is negative, the bisector corresponding to the acute angle is given by taking the positive sign on the right side of the formula (i.e., using \(+\)).
\(\frac{3x - 4y + 7}{5} = + \frac{-12x - 5y + 2}{13}\)
Cross-multiply:
\(13(3x - 4y + 7) = 5(-12x - 5y + 2)\)
\(39x - 52y + 91 = -60x - 25y + 10\)
Move all terms to one side:
\(39x + 60x - 52y + 25y + 91 - 10 = 0\)
\(99x - 27y + 81 = 0\)
Divide the entire equation by 9 to simplify:
\(11x - 3y + 9 = 0\) ... (3)
This is the equation of the bisector of the acute angle.
(For completeness, let's find the obtuse angle bisector by taking the negative sign:
\(\frac{3x - 4y + 7}{5} = - \frac{-12x - 5y + 2}{13}\)
\(13(3x - 4y + 7) = -5(-12x - 5y + 2)\)
\(39x - 52y + 91 = 60x + 25y - 10\)
\(39x - 60x - 52y - 25y + 91 + 10 = 0\)
\(-21x - 77y + 101 = 0\)
\(21x + 77y - 101 = 0\) ... (4)
This is the obtuse angle bisector.)
To confirm the acute angle choice, we can use the angle formula between a line and an angle bisector. The angle \(\theta\) between the acute angle bisector (3) and either line (1) or (2) should be less than \(45^\circ\).
Let's check with line (1):
Slope of line (3) \(m_3 = -\frac{11}{-3} = \frac{11}{3}\)
Slope of line (1) \(m_1 = -\frac{3}{-4} = \frac{3}{4}\)
\(\tan \theta = \left|\frac{m_3 - m_1}{1 + m_3m_1}\right| = \left|\frac{\frac{11}{3} - \frac{3}{4}}{1 + \frac{11}{3} \times \frac{3}{4}}\right|\)
\(\tan \theta = \left|\frac{\frac{44 - 9}{12}}{1 + \frac{33}{12}}\right| = \left|\frac{\frac{35}{12}}{\frac{12+33}{12}}\right| = \left|\frac{\frac{35}{12}}{\frac{45}{12}}\right| = \frac{35}{45} = \frac{7}{9}\)
Since \(\tan \theta = \frac{7}{9} < 1\), it means \(\theta < 45^\circ\), confirming that \(11x - 3y + 9 = 0\) is indeed the bisector of the acute angle. The acute angle bisector helps to divide the space into two parts with smaller angles.
In simple words: We have two lines, and we want to find the line that cuts the *smaller* angle between them exactly in half. We used a formula that gives two possible bisector lines. To pick the correct one, we looked at a special calculation involving the numbers in front of \(x\) and \(y\) from the original lines. If this calculation is negative, we use the positive sign in the formula for the acute angle. This led us to the equation \(11x - 3y + 9 = 0\). We also double-checked this by finding the steepness of this bisector line and confirming it makes a small angle with one of the original lines.

🎯 Exam Tip: To identify the acute/obtuse angle bisector, remember the condition \(A_1A_2 + B_1B_2\). If \(A_1A_2 + B_1B_2 > 0\), then \(+\) sign gives the obtuse angle bisector and \(-\) sign gives the acute. If \(A_1A_2 + B_1B_2 < 0\), then \(+\) sign gives the acute angle bisector and \(-\) sign gives the obtuse. Always ensure the constant terms \(C_1\) and \(C_2\) are positive before applying this rule.