JEE Mathematics Straight Lines MCQs Set 12

Answer: (d) \( -\tfrac35x+\tfrac45y=1 \)

Explanation:
1. Write 3x - 4y = -5, then multiply by -1 to make the right side positive: -3x + 4y = 5.
2. Divide by sqrt(9 + 16) = 5: \( -\tfrac35x+\tfrac45y=1 \).
3. So p = 1, cos α = -3/5, sin α = 4/5.

Teacher's Note:
1. In normal form p must be positive.
2. Divide by sqrt(a² + b²) after fixing the sign.

Answer: (a) x = 3

Explanation:
1. The opposite side joins (0, 0) and (6, 0) with mid-point (3, 0).
2. The median through (3, 8) and (3, 0) is the vertical line x = 3.

Teacher's Note:
1. A median joins a vertex to the mid-point of the opposite side.
2. Equal x-coordinates mean a vertical line.

Answer: (b) 8x - 9y = 0

Explanation:
1. The opposite side joins (6, 0) and (3, 8), with mid-point (9/2, 4).
2. Slope = 4/(9/2) = 8/9, so the median is y = 8x/9, i.e. 8x - 9y = 0.

Teacher's Note:
1. Find the mid-point of the opposite side first.
2. Then use the two-point form.

Answer: (c) \( \frac1{a^2}+\frac1{b^2}=\frac1{p^2} \)

Explanation:
1. The line is x/a + y/b = 1, i.e. bx + ay - ab = 0.
2. \( p=\frac{ab}{\sqrt{a^2+b^2}}\Rightarrow\frac1{p^2}=\frac{a^2+b^2}{a^2b^2}=\frac1{a^2}+\frac1{b^2} \)

Teacher's Note:
1. Use the perpendicular distance formula from the origin.
2. This relation is also the pedal form of a tangent to an astroid.

Answer: (a) \( \text{square of side } 2\sqrt2\text{ units} \)

Explanation:
1. The boundary lines are x - y = ±2 and x + y = ±2.
2. Their vertices are (2, 0), (0, 2), (-2, 0), (0, -2); sides are of length \( \sqrt{4+4}=2\sqrt2 \) and adjacent sides are perpendicular.
3. So it is a square of side \( 2\sqrt2 \) (area 8).

Teacher's Note:
1. Lines x - y = c and x + y = d are mutually perpendicular.
2. Find the vertices and check side lengths and angles.

Answer: (b) 3

Explanation:
1. Line x/a + y/b = 1 with 2/a + 4/b = 1 and |ab|/2 = 16, i.e. ab = ±32.
2. ab = 32: 2b + 4a = 32 with ab = 32 gives (a - 4)² = 0, so a = 4, b = 8 (one line).
3. ab = -32: 2b + 4a = -32 with ab = -32 gives a² + 8a - 16 = 0, with two real roots, hence two lines.
4. Total = 3 lines.

Teacher's Note:
1. Treat the two signs of the product of intercepts separately.
2. Count only real solutions of each quadratic.

Answer: (c) 3x - 2y = 12

Explanation:
1. x/4 + y/b = 1 through (2, -3): 1/2 - 3/b = 1, so b = -6.
2. x/4 - y/6 = 1 gives 3x - 2y = 12.

Teacher's Note:
1. Use the double-intercept form and solve for the unknown intercept.
2. Check: 3(4) - 0 = 12.

Answer: (d) x + y = 7

Explanation:
1. x/a + y/a = 1 means x + y = a.
2. (2, 5) gives a = 7, so x + y = 7.

Teacher's Note:
1. Equal intercepts give slope -1.
2. Substitute the given point.

Answer: (a) x + 4y = 8

Explanation:
1. x/a + y/b = 1 with 4/a + 1/b = 1 and ab = 16.
2. Then 4b + a = ab = 16 gives 4b² - 16b + 16 = 0, so b = 2 and a = 8.
3. The line is x/8 + y/2 = 1, i.e. x + 4y = 8.

Teacher's Note:
1. Use both the passing-through condition and the area condition.
2. The quadratic has a repeated root, so the line is unique.

Answer: (b) 780

Explanation:
1. Integer points (x, y) with x ≥ 1, y ≥ 1 and x + y ≤ 40.
2. For x = 1, 2, ..., 39 the number of y values is 40 - x, so the total is 39 + 38 + ... + 1.
3. Total = 39 × 40 / 2 = 780.

Teacher's Note:
1. Interior points need x ≥ 1, y ≥ 1 and x + y < 41.
2. Sum an arithmetic series.

Answer: (c) 18

Explanation:
1. Line x/a + y/b = 1 with 8/a + 2/b = 1. Then b = 2a/(a - 8) and a + b = a + 2a/(a - 8).
2. By Cauchy-Schwarz, \( (a+b)\left(\tfrac8a+\tfrac2b\right)\ge(\sqrt8+\sqrt2)^2=18 \), with equality attainable (a = 12, b = 6).
3. So the minimum of OP + OQ is 18.

Teacher's Note:
1. Minimise a + b subject to 8/a + 2/b = 1.
2. Cauchy-Schwarz or calculus gives the same minimum.

Answer: (d) x - 6y = 0 and 2x - 3y = 0

Explanation:
1. The line meets the axes at (12, 0) and (0, 4).
2. Trisection points: (8, 4/3) and (4, 8/3).
3. Lines through the origin: y = x/6, i.e. x - 6y = 0, and y = 2x/3, i.e. 2x - 3y = 0.

Teacher's Note:
1. Find the axis intercepts, then the two trisection points.
2. A line through the origin has equation y = mx.

Answer: (a) 5

Explanation:
1. Slope of 2x - 3y + 17 = 0 is 2/3, so the other slope is -3/2.
2. (β - 17)/(15 - 7) = -3/2 gives β - 17 = -12, so β = 5.

Teacher's Note:
1. Perpendicular slopes multiply to -1.
2. Solve the linear equation for β.

Answer: (b) 12x - 8y = 7

Explanation:
1. A = (7/2, 0) and B = (0, 7/3); the mid-point is (7/4, 7/6).
2. Slope of AB is -2/3, so the perpendicular bisector has slope 3/2.
3. y - 7/6 = (3/2)(x - 7/4), which simplifies to 12x - 8y = 7.

Teacher's Note:
1. A perpendicular bisector passes through the mid-point with the negative reciprocal slope.
2. Check that the mid-point satisfies the final equation: 21 - 28/3·... = 7.

Answer: (c) x - y = 0

Explanation:
1. The sides meet at O(0, 0), which does not lie on 11x + 7y = 9, so that diagonal joins the other two vertices.
2. It meets 4x + 5y = 0 at (5/3, -4/3) and 7x + 2y = 0 at (-2/3, 7/3).
3. The other diagonal passes through O and the mid-point (1/2, 1/2) of these vertices: y = x.

Teacher's Note:
1. The two diagonals bisect each other.
2. The diagonal through the common vertex O passes through the mid-point of the other diagonal.

Answer: (a) 2x + 3y + 5 = 0

Explanation:
1. A parallel line has the same x and y coefficients: 2x + 3y = k.
2. Through (2, -3): 4 - 9 = k = -5, so 2x + 3y + 5 = 0.

Teacher's Note:
1. Parallel lines differ only in the constant term.
2. Substitute the given point.

Answer: (b) 3x - 2y = 12

Explanation:
1. A perpendicular line has the form 3x - 2y = k.
2. Through (2, -3): 6 + 6 = 12, so 3x - 2y = 12.

Teacher's Note:
1. Swap the coefficients and change one sign for the perpendicular.
2. Substitute the point to find the constant.

Answer: (c) (2, 6) and (3, 7)

Explanation:
1. Altitude from A is perpendicular to BC (slope 1), so it has slope -1: x + y = 8. Altitude from B is perpendicular to AC (slope 0), so x = 2. They meet at (2, 6).
2. Circumcentre: equidistant from the vertices; perpendicular bisector of AC is x = 3, and equating distances to A and B gives y = 7: (3, 7).
3. (The centroid (8/3, 20/3) lies on the line joining them in the ratio 2 : 1 as expected.)

Teacher's Note:
1. Orthocentre: intersection of altitudes. Circumcentre: intersection of perpendicular bisectors.
2. Euler line check: G divides HO in the ratio 2 : 1.

Answer: (d) \( -3,\ \tfrac12,\ 1 \)

Explanation:
1. Slopes: -2, 1 and 3. The vertices are (-2/3, 13/3), (2, 7) and (2/5, 11/5).
2. Using the vertices, tan of the interior angles come out as -3 (obtuse), 1/2 and 1.
3. Check: tan A + tan B + tan C = -3/2 = tan A tan B tan C = -3/2.

Teacher's Note:
1. In any triangle tan A + tan B + tan C = tan A tan B tan C, a useful check.
2. A negative tangent indicates an obtuse interior angle.

Answer: (a) x - 5y + 9 = 0 or 5x + y - 7 = 0

Explanation:
1. Slope of the given line is -2/3. For a line of slope m at 45°: \( \left|\frac{m+\frac23}{1-\frac23m}\right|=1 \).
2. This gives m = 1/5 or m = -5.
3. Through (1, 2): y - 2 = (x - 1)/5, i.e. x - 5y + 9 = 0, and y - 2 = -5(x - 1), i.e. 5x + y - 7 = 0.

Teacher's Note:
1. There are always two lines making a given angle with a given line.
2. Use tan θ = |(m1 - m2)/(1 + m1 m2)|.

Answer: (b) x + 2 = 0 and 7x + 24y + 182 = 0

Explanation:
1. Slope of the given line is -4/3, and tan θ = 3/4 (the line is at an angle whose tangent is 3/4).
2. Let the required slope be m: \( \left|\frac{m+\frac43}{1-\frac43m}\right|=\frac34 \). This gives m = -7/24, and the other line is vertical (the denominator vanishes).
3. Vertical line through (-2, -7): x + 2 = 0. Slope -7/24: y + 7 = -(7/24)(x + 2), i.e. 7x + 24y + 182 = 0.

Teacher's Note:
1. Do not forget the vertical line, which is missed when slopes alone are used.
2. A vertical line makes angle with slope -4/3: tan = 3/4.

Answer: (d) \( \left(\tfrac{11}{5},\tfrac25\right) \)

Explanation:
1. ax1 + by1 + c = 1 - 4 = -3 and a² + b² = 5.
2. x' = 1 - 2(1)(-3)/5 = 11/5 and y' = -2 - 2(2)(-3)/5 = 2/5.

Teacher's Note:
1. Use the standard image formula and check the mid-point lies on the line.
2. Mid-point (8/5, -4/5): 8/5 - 8/5 = 0 on the line.

Answer: (c) 31/10

Explanation:
1. Rewrite the second as 3x + 4y - 17/2 = 0.
2. Distance = \( \frac{|7-(-17/2)|}{5}=\frac{31/2}{5}=\frac{31}{10} \).

Teacher's Note:
1. Make the x and y coefficients equal before using the formula.
2. Distance between ax + by + c1 = 0 and ax + by + c2 = 0 is |c1 - c2|/sqrt(a² + b²).

Answer: (d) \( \left(-\tfrac14,\tfrac23\right) \)

Explanation:
1. The lines are 4x - 3y + c = 0 with |c|/5 = 3/5, so c = ±3.
2. Test (-1/4, 2/3): 4(-1/4) - 3(2/3) + 3 = -1 - 2 + 3 = 0, so it lies on 4x - 3y + 3 = 0.
3. The other points give 4x - 3y = 0, 1 and 2, none equal to ±3 (need 4x - 3y = ∓3).

Teacher's Note:
1. A line parallel to the given one has the same x and y coefficients.
2. Substitute each option in 4x - 3y = ±3.