# CBSE Class 11th Math 10 – Straight Lines MCQs

#### The equation of the line whose sum of the intercepts is 3 and their product is 2, a > b, is ____.

Given a + b = 3; ab = 2; a > b ⇒ a = 2; b = 1 Equation of the line is (x/a) + (y/b) = 1 ⇒ (x/2) + (y/1) = 1 ⇒ x + 2y - 2 = 0

#### The equation of the line that cuts off an intercept of 4 units in the negative direction of the Y axis and makes an angle 135° with the positive direction of X axis is ____.

The equation of the line is given by y = mx + c, where m = tan θ = tan 135° = tan (90 + 45) = -1 c = y intercept = -4 Therefore, y = (-1)(x) + (-4) = -x – 4 ⇒ x + y + 4 = 0

#### If the general equation of a straight line is 2x + 3y − 9 = 0, then the y-intercept of the line is ____.

The general equation of a straight line is 2x + 3y −9 = 0. It can be written as 3y = −2x + 9. And hence, y = (−2/3)x + 3 - - - - (1) Equation (1) is in the form of the slope−intercept form, y = mx + c, where c is the y-intercept. Therefore, the y-intercept of the given straight line is 3.

#### If the general equation of two parallel straight lines are x − my + 6 = 0 and −mx + y − 1 = 0, then the value of m is ____.

The general equation of two straight lines are a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0, where b_{1}, b_{2} ≠ 0. They are parallel if a_{1/}b_{1}= a_{2/}b_{2}. The given general equation of the two straight lines, x – my + 6 = 0 and −mx + y − 1 = 0, where a1 = 1 a_{2} = −m, b_{1 }= −m and b_{2} = 1. So, 1/−m = −m/1, ⇒ m^{2} = 1

#### The equation of the line making an angle Π/4 with the X axis and passing through (2, 1) is ____.

The equation of the line is given by, y - y_{1} = m (x - x_{1}) where m = tan θ. m = tan Π/4 = 1 The line passes through (2, 1). Therefore, (x_{1}, y_{1}) is (2, 1). ⇒ y - 1 = 1(x - 2) ⇒ y - 1 = x – 2 ⇒ x - y - 1 = 0

#### If 3 and −2 are the x-intercept and y-intercept of a straight line respectively, and then the general equation of the straight line is ____.

If 3 and −2 are the x-intercept and y-intercept of a straight line respectively, then the equation of the straight line is x/3 + y/(−2) = 1 ⇒ 2x – 3y = 6 ⇒ 2x – 3y – 6 = 0 (which is in the form of Ax + By + C = 0, i.e. general equation of the straight line) Hence, the general equation of the straight line is 2x – 3y – 6 = 0.

#### Find the slope of the line perpendicular to the line AB, if A is (3, 3) and B is (-1, 1).

Let the slope of AB be m. Slope of the line perpendicular to the line AB = -1/m. Slope of AB = (y_{2 }- y_{1})/(x_{2 }- x_{1}) = 1-3 /-1-3 = 12 Slope of the line perpendicular to AB is -1/(1/2) = -2

#### If the general equation of a straight line is x − y − 6 = 0, then the slope of the line is ____.

The general equation of a straight line is x − y −6 = 0. It can be written as −y = −x + 6 And hence, y = x − 6…(1) Equation (1) is in form of a slope-intercept form y = mx + c, where m is the slope of the straight line. Therefore, slope of the given straight line is 1.

#### The equation of the straight line cutting off an intercept -2 from Y axis and equally inclined to the axes is ____.

Given C = -2. Equal inclination ⇒ θ = 45° ⇒ slope (m) = tan 45° = 1 or m = tan 135° = -1 Substituting these values in y = mx + c, we get, y = (1)x + (-2) or y = (-1)x + (-2) y = x - 2 or y = -x - 2 ⇒ x - y - 2 = 0 or x + y + 2 = 0 Of the options given, x + y + 2 = 0 is the right answer.

#### If the slope of AB = slope of CD and AB = CD, then the quadrilateral ABCD is a ____.

As per the properties of a parallelogram, the slopes of opposite sides are equal.

#### The equation of the line CD parallel to the line joining A(-2, 3) and B(-4, 2) and passing through the point (1, 0) is ____ .

Slope of CD = Slope of AB = m (because the lines are parallel) m = (y_{2 }- y_{1})/(x_{2 }- x_{1}) = (- 4 + 2) /(2 – 3) = 2 The equation of the line is y - y_{1 }= m (x - x_{1}). (1, 0) lies on the line. Therefore, y - 0 = 2(x - 1) ⇒ 2x - y + 2 = 0

#### If the general equations of the two straight lines are 2x − 3y + 6 = 0 and −5x + λy − 1 = 0 and they are perpendicular each other, then the value of λ is ____.

The general equations of the two straight lines are a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0, where b_{1}, b_{2} ≠ 0. They are perpendicular if a_{1}a_{2} + b_{1}b_{2} = 0. The given general equations of the two straight lines, 2x − 3y + 6 = 0 and −5x + λy − 1 = 0, where a_{1} = 2, a_{2}= −5, b_{1} = −3 and b_{2} = λ. So, 2 × (−5) + (−3) × λ = 0 λ = 10/3 Hence, the value of λ is 10/3.

#### If (3, 4) is the mid point of the line segment intercepted between the axes, the equation of the line is ____.

Equation of the line is x/a + y/b = 0 (3, 4) is the midpoint of AB Given, a/2 = 3 ⇒ a = 6 b/2 = 4 ⇒ b = 8 x/6 + y/8 = 1 ⇒ 3x + 4y = 24

#### The equation of the line that cuts off an intercept 2 on the positive direction of X axis and an intercept 4 on the negative direction of Y axis is ____.

Given a = 2; b = -4 Equation of the line is x/a + y/ b = 1 ⇒ (x/2) + (y/-4) = 1 ⇒ 2x - y = - 4 ⇒2x - y + 4 = 0

#### 2x + 2y - 1 = 0 and -6x - 6y + 3 = 0 are ____.

From the 2 equations, we find that a_{1/}a_{2 }= b_{1/}b_{2 }= c_{1/}c_{2 } = -1/3 Therefore, the lines are coincident.

#### If the line joining the points A (a, -a) and B (2, -3) is parallel to the line 3x + 4y - 5 = 0, then the value of 'a' is ____.

Since both lines are parallel, the slopes of the two lines are the same. Slope of AB = (y_{2 }- y_{1})/(x_{2 }- x_{1}) = −3+a /2−a 3x + 4y - 5 = 0 ⇒ y = (−3/4)x + (5/4) Hence, slope of 3x + 4y - 5 = 0 is −3/4. ∴(−3+a )/(2−a) = −34 ⇒ −12 + 4a = −6 + 3a ∴ a = 6

#### The equation of the straight line, whose perpendicular length from the origin is 13 units and the slope of the perpendicular is 5/12, ____.

p = 13, tan∝ = 5/12 ⇒ sin∝ = 5/13 and cos∝ = 12/13 The equation of the line is given by is x cos α + y sin α = p. x? 12/13 + y? 5/13 = 13 ⇒12x + 5y = 169 ⇒12x + 5y - 169 = 0

#### The equation of the line passing through (-3, 5) and perpendicular to the line x - 2y + 3 = 0 is ____.

Given equation is x - 2y + 3 = 0 ⇒y =( 1/2)x + (3/2) ⇒ m_{1}= 1/2 The slope of the required line, m_{2}= -1/m_{1} (because the lines are perpendicular) ⇒ -1/(1/2) = - 2 The equation of the line with the slope m = -2, and passing through (-3, 5) is given by, y - y_{1} = m(x - x_{1}) (y - 5) = (-2) (x + 3) y - 5 = -2x – 6 ⇒2x + y + 1 = 0

#### If the line (a - 5)x + (a^{2} - 2)y + (a - 1)(a - 5) = 0 passes through the origin, then the value of a is ____ .

Since the line passes through the origin, substitute x = 0 and y = 0 in the given equation. (a - 5)0 + (a^{2} - 2)0 + (a - 1)(a - 5) = 0 ⇒ 0 + 0 + (a - 1)(a - 5) = 0 ⇒ (a - 1)(a - 5) = 0 ⇒ a - 1 = 0 or a - 5 = 0 ⇒ a = 1 or a = 5

#### Choose the option that is false.

If 90° < θ < 180°, then θ is obtuse and slope is negative.

#### If (a, 0), (0, b) and (x, y) are collinear, using the concept of slope, we get the equation ____.

If the three points A, B and C are collinear, slope of BC with a common point B Slope of AB = (b – 0)/(0 – a) = −b/a Slope of BC = (y − b)/(x−0) = y−b/x Slope of AB = Slope of BC −b/a = y-b/x ⇒ −bx = ay – ab ⇒ bx + ay = ab

#### The equation of the line passing through (3, -2) and having the y intercept twice that of the x intercept is ____.

Given b = 2a ⇒ x/a + y/2a = 1 ⇒2x + y = 2a It passes through (3, -2) ⇒ 2(3) + (-2) = 2a ⇒ 4 = 2a Therefore, the equation of the line is 2x + y = 4 (because 2a = 4)

#### The slope of the line parallel to the line joining the points (5, +1) and (-3, 9) is ____.

Slope=( y_{2 }- y_{1})/(x_{2 }- x_{1}) = (9-1)/(-3 – 5) = 8 – 8 = -1 Slope of the line parallel to this line is also -1, as slopes of parallel lines are the same.

#### The angle between the two lines x + y - 5 = 0 and y = x + 3 is ____.

Consider the equation x + y - 5 = 0. Comparing with y = mx + c, we get slope m_{1} = -1 For the equation y = x + 3, we get slope m_{2} = 1 Since, the product of the slopes m_{1}m_{2} = -1, the lines are perpendicular to each other. Hence, the angle between the lines is 90° or π/2.

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