Maharashtra Board Class 10 Maths Chapter 5 Co ordinate Geometry Set 5.3 Solutions

Question 1. Angles made by the line with the positive direction of X-axis are given. Find the slope of these lines.
(i) 45°
(ii) 60°
(iii) 90°
Answer:
(i) Angle made with the positive direction of X-axis \((\theta) = 45^\circ\)
Slope of the line \((\text{m}) = \tan \theta\)
\( \implies \text{m} = \tan 45^\circ = 1\)
\( \therefore \) The slope of the line is 1.

(ii) Angle made with the positive direction of X-axis \((\theta) = 60^\circ\)
Slope of the line \((\text{m}) = \tan \theta\)
\( \implies \text{m} = \tan 60^\circ = \sqrt{3}\)
\( \therefore \) The slope of the line is \(\sqrt{3}\).

(iii) Angle made with the positive direction of X-axis \((\theta) = 90^\circ\)
Slope of the line \((\text{m}) = \tan \theta\)
\( \implies \text{m} = \tan 90^\circ\)
But, the value of \(\tan 90^\circ\) is not defined.
\( \therefore \) The slope of the line cannot be determined.
In simple words: The slope of a line is determined by the tangent of the angle it makes with the positive X-axis. For angles 45° and 60°, the slopes are 1 and \(\sqrt{3}\) respectively, but for 90°, the slope is undefined as tan 90° is undefined.

🎯 Exam Tip: Remember that the slope (m) of a line is given by \( \text{m} = \tan \theta \), where \( \theta \) is the angle the line makes with the positive X-axis. Be mindful of undefined tangent values for specific angles like 90°. The values of standard trigonometric angles are crucial here.

Question 2. Find the slopes of the lines passing through the given points.
(i) A (2, 3), B (4, 7)
(ii) P(-3, 1), Q (5, -2)
(iii) C (5, -2), D (7, 3)
(iv) L (-2, -3), M (-6, -8)
(v) E (-4, -2), F (6, 3)
(vi) T (0, -3), S (0,4)
Answer:
(i) A (x\(_1\), y\(_1\)) = A (2, 3) and B (x\(_2\), y\(_2\)) = B (4, 7)
Here, x\(_1\) = 2, x\(_2\) = 4, y\(_1\) = 3, y\(_2\) = 7
Slope of line AB = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{7-3}{4-2} = \frac{4}{2} = 2 \)
\( \therefore \) The slope of line AB is 2.

(ii) P (x\(_1\), y\(_1\)) = P (-3, 1) and Q (x\(_2\), y\(_2\)) = Q (5, -2)
Here, x\(_1\) = -3, x\(_2\) = 5, y\(_1\) = 1, y\(_2\) = -2
Slope of line PQ = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{-2-1}{5-(-3)} = \frac{-3}{5+3} = \frac{-3}{8} \)
\( \therefore \) The slope of line PQ is \( -\frac{3}{8} \).

(iii) C (x\(_1\), y\(_1\)) = C (5, -2) and D (x\(_2\), y\(_2\)) = D (7, 3)
Here, x\(_1\) = 5, x\(_2\) = 7, y\(_1\) = -2, y\(_2\) = 3
Slope of line CD = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{3-(-2)}{7-5} = \frac{3+2}{2} = \frac{5}{2} \)
\( \therefore \) The slope of line CD is \( \frac{5}{2} \).

(iv) L (x\(_1\), y\(_1\)) = L (-2, -3) and M (x\(_2\),y\(_2\)) = M (-6, -8)
Here, x\(_1\) = -2, x\(_2\) = -6, y\(_1\) = -3, y\(_2\) = -8
Slope of line LM = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{-8-(-3)}{-6-(-2)} = \frac{-8+3}{-6+2} \)
\( \implies = \frac{-5}{-4} = \frac{5}{4} \)
\( \therefore \) The slope of line LM is \( \frac{5}{4} \).

(v) E (x\(_1\), y\(_1\)) = E (-4, -2) and F (x\(_2\), y\(_2\)) = F (6, 3)
Here,x\(_1\) = -4, x\(_2\) = 6, y\(_1\) = -2, y\(_2\) = 3
Slope of line EF = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{3-(-2)}{6-(-4)} = \frac{3+2}{6+4} \)
\( \implies = \frac{5}{10} = \frac{1}{2} \)
\( \therefore \) The slope of line EF is \( \frac{1}{2} \).

(vi) T (x\(_1\), y\(_1\)) = T (0, -3) and S (x\(_2\), y\(_2\)) = S (0, 4)
Here, x\(_1\) = 0, x\(_2\) = 0, y\(_1\) = -3, y\(_2\) = 4
Slope of line TS = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{4-(-3)}{0-0} = \frac{4+3}{0} = \frac{7}{0} \)
\( \implies = \) Not defined
\( \therefore \) The slope of line TS cannot be determined.
In simple words: The slope of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is calculated using the formula \( \frac{y_2-y_1}{x_2-x_1} \). If the x-coordinates are the same (a vertical line), the slope is undefined.

🎯 Exam Tip: Master the slope formula \( \text{m} = \frac{y_2-y_1}{x_2-x_1} \). Pay close attention to negative signs and subtraction, as small errors can lead to incorrect slopes. Remember that a vertical line has an undefined slope.

Question 3. Determine whether the following points are collinear.
(i) A (-1, -1), B (0, 1), C (1, 3)
(ii) D (- 2, -3), E (1, 0), F (2, 1)
(iii) L (2, 5), M (3, 3), N (5, 1)
(iv) P (2, -5), Q (1, -3), R (-2, 3)
(v) R (1, -4), S (-2, 2), T (-3,4)
(vi) A(-4,4),K[-2,\(\frac{5}{2}\)], N (4,-2)
Answer:
(i) Slope of line AB = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{1-(-1)}{0-(-1)} = \frac{1+1}{0+1} = 2 \)
Slope of line BC = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{3-1}{1-0} = 2 \)
\( \therefore \) slope of line AB = slope of line BC
\( \therefore \) line AB || line BC
Also, point B is common to both the lines.
\( \therefore \) Both lines are the same.
\( \therefore \) Points A, B and C are collinear.

(ii) Slope of line DE = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{0-(-3)}{1-(-2)} \)
\( \implies = \frac{0+3}{1+2} = \frac{3}{3} = 1 \)
Slope of line EF = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{1-0}{2-1} = 1 \)
\( \therefore \) slope of line DE = slope of line EF
\( \therefore \) line DE || line EF
Also, point E is common to both the lines.
\( \therefore \) Both lines are the same.
\( \therefore \) Points D, E and F are collinear.

(iii) Slope of line LM = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{3-5}{3-2} = \frac{-2}{1} = -2 \)
Slope of line MN = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{1-3}{5-3} = \frac{-2}{2} = -1 \)
\( \therefore \) slope of line LM \( \ne \) slope of line MN
\( \therefore \) Points L, M and N are not collinear.

(iv) Slope of line PQ = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{-3-(-5)}{1-2} \)
\( \implies = \frac{-3+5}{-1} = \frac{2}{-1} = -2 \)
Slope of line QR = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{3-(-3)}{-2-1} \)
\( \implies = \frac{3+3}{-3} = \frac{6}{-3} = -2 \)
\( \therefore \) slope of line PQ = slope of line QR
\( \therefore \) line PQ || line QR
Also, point Q is common to both the lines.
\( \therefore \) Both lines are the same.
\( \therefore \) Points P, Q and R are collinear.

(v) Slope of line RS = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{2-(-4)}{-2-1} \)
\( \implies = \frac{2+4}{-3} = \frac{6}{-3} = -2 \)
Slope of line ST = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{4-2}{-3-(-2)} \)
\( \implies = \frac{2}{-3+2} = \frac{2}{-1} = -2 \)
\( \therefore \) slope of line RS = slope of line ST
\( \therefore \) line RS || line ST
Also, point S is common to both the lines.
\( \therefore \) Both lines are the same.
\( \therefore \) Points R, S and T are collinear.

(vi) Slope of line AK = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{\frac{5}{2}-4}{-2-(-4)} = \frac{\frac{5-8}{2}}{-2+4} = \frac{-\frac{3}{2}}{2} \)
\( \implies = -\frac{3}{4} \)
Slope of line KN = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{-2-\frac{5}{2}}{4-(-2)} = \frac{\frac{-4-5}{2}}{4+2} \)
\( \implies = \frac{-\frac{9}{2}}{6} = -\frac{9}{12} = -\frac{3}{4} \)
\( \therefore \) slope of line AK = slope of line KN
\( \therefore \) line AK || line KN
Also, point K is common to both the lines.
\( \therefore \) Both lines are the same.
\( \therefore \) Points A, K and N are collinear.
In simple words: To check if three or more points are collinear, calculate the slopes of line segments formed by consecutive pairs of points. If the slopes are equal and the line segments share a common point, then the points are collinear.

🎯 Exam Tip: For collinearity questions, remember the key principle: if three points A, B, and C are collinear, then the slope of AB must equal the slope of BC. Always mention the common point and parallel lines in your explanation for full marks.

Question 4. If A (1, -1), B (0,4), C (-5,3) are vertices of a triangle, then find the slope of each side.
Answer:
We know that, slope of line = \( \frac{y_2-y_1}{x_2-x_1} \)
Slope of side AB = \( \frac{4-(-1)}{0-1} = \frac{4+1}{-1} = \frac{5}{-1} = -5 \)
Slope of side BC = \( \frac{3-4}{-5-0} = \frac{-1}{-5} = \frac{1}{5} \)
Slope of side AC = \( \frac{3-(-1)}{-5-1} = \frac{3+1}{-6} = \frac{4}{-6} = -\frac{2}{3} \)
\( \therefore \) The slopes of the sides AB, BC and AC are -5, \( \frac{1}{5} \) and \( -\frac{2}{3} \) respectively.
In simple words: To find the slope of each side of a triangle given its vertices, apply the slope formula \( \frac{y_2-y_1}{x_2-x_1} \) to each pair of vertices that form a side (AB, BC, and AC).

🎯 Exam Tip: Be careful with the coordinates and the order of subtraction when calculating slopes. A common mistake is flipping the signs or mixing up x and y values. Double-check your calculations, especially with negative coordinates.

Question 5. Show that A (-4, -7), B (-1, 2), C (8, 5) and D (5, -4) are the vertices of a parallelogram.
Answer:
Proof:
We know that, slope of line = \( \frac{y_2-y_1}{x_2-x_1} \)
Slope of side AB = \( \frac{2-(-7)}{-1-(-4)} \)
\( \implies = \frac{2+7}{-1+4} = \frac{9}{3} = 3 \) ...(i)
Slope of side BC = \( \frac{5-2}{8-(-1)} \)
\( \implies = \frac{3}{8+1} = \frac{3}{9} = \frac{1}{3} \) ...(ii)
Slope of side CD = \( \frac{-4-5}{5-8} \)
\( \implies = \frac{-9}{-3} = 3 \) ...(iii)
Slope of side AD = \( \frac{-4-(-7)}{5-(-4)} \)
\( \implies = \frac{-4+7}{5+4} = \frac{3}{9} = \frac{1}{3} \) ...(iv)
\( \therefore \) Slope of side AB = Slope of side CD ... [From (i) and (iii)]
\( \therefore \) side AB || side CD
Slope of side BC = Slope of side AD ... [From (ii) and (iv)]
\( \therefore \) side BC || side AD
Both the pairs of opposite sides of \(\text{ABCD}\) are parallel.
\(\text{ABCD}\) is a parallelogram.
Points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a parallelogram.
In simple words: To prove that a quadrilateral is a parallelogram, we show that both pairs of opposite sides are parallel. This is done by calculating the slopes of all four sides and demonstrating that the slopes of opposite sides are equal.

🎯 Exam Tip: The key to proving a figure is a parallelogram using slopes is to show that `Slope(AB) = Slope(CD)` and `Slope(BC) = Slope(AD)`. This verifies that opposite sides are parallel. Clearly label each slope calculation and state your conclusion based on the slope equalities.

Question 6. Find k, if R (1, -1), S (-2, k) and slope of line RS is -2.
Answer:
R(x\(_1\), y\(_1\)) = R (1, -1), S (x\(_2\), y\(_2\)) = S (-2, k)
Here, x\(_1\) = 1, x\(_2\) = -2, y\(_1\) = -1, y\(_2\) = k
Slope of line RS = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{k-(-1)}{-2-1} = \frac{k+1}{-3} \)
But, slope of line RS is -2. ... [Given]
\( \therefore -2 = \frac{k+1}{-3} \)
\( \therefore k + 1 = 6 \)
\( \therefore k = 6-1 \)
\( \therefore k = 5 \)
In simple words: Given two points and the slope of the line connecting them, we can use the slope formula \( \text{m} = \frac{y_2-y_1}{x_2-x_1} \) to set up an equation and solve for the unknown coordinate 'k'.

🎯 Exam Tip: When an unknown variable is involved in coordinate geometry problems, carefully substitute the given values into the appropriate formula. Algebraic manipulation (like cross-multiplication) is often required to solve for the unknown, so be precise with your steps.

Question 7. Find k, if B (k, -5), C (1, 2) and slope of the line is 7.
Answer:
B(x\(_1\), y\(_1\)) = B (k, -5), C (x\(_2\), y\(_2\)) = C (1, 2)
Here, x\(_1\) = k, x\(_2\) = 1, y\(_1\) = -5, y\(_2\) = 2
Slope of line BC = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{2-(-5)}{1-k} = \frac{2+5}{1-k} \)
\( \implies = \frac{7}{1-k} \)
But, slope of line BC is 7. ...[Given]
\( \therefore 7 = \frac{7}{1-k} \)
\( \therefore 7(1 - k) = 7 \)
\( \therefore 1-k = \frac{7}{7} \)
\( \therefore 1 - k = 1 \)
\( \therefore k = 0 \)
In simple words: Given the coordinates of two points with one unknown (k) and the slope of the line, substitute these values into the slope formula to form an equation. Solve this equation for 'k' to find its value.

🎯 Exam Tip: Always write down the given information clearly. When solving equations involving fractions, cross-multiplication is a useful technique. Simplify carefully to avoid errors in the final value of 'k'.

Question 8. Find k, if PQ || RS and P (2, 4), Q (3, 6), R (3,1), S (5, k).
Answer:
Slope of line PQ = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{6-4}{3-2} = \frac{2}{1} = 2 \)
Slope of line RS = \( \frac{y_2-y_1}{x_2-x_1} \)
\( \implies = \frac{k-1}{5-3} = \frac{k-1}{2} \)
But, line PQ || line RS ... [Given]
\( \therefore \) Slope of line PQ = Slope of line RS
\( \therefore 2 = \frac{k-1}{2} \)
\( \therefore 4 = k-1 \)
\( \therefore k = 4 + 1 \)
\( \therefore k = 5 \)
In simple words: If two lines are parallel, their slopes are equal. First, calculate the slope of the line with known coordinates (PQ). Then, set this slope equal to the slope of the line with an unknown coordinate (RS) and solve for 'k'.

🎯 Exam Tip: The core concept here is that parallel lines have equal slopes. This property is fundamental for solving problems where collinearity or parallelism is given. Ensure accurate calculation of both slopes before equating them and solving for the unknown.