Straight Lines- Class X ICSE Maths MCQs

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Every straight line can be represented by a linear equation.

The equations 2x+5y=1, x-3y = 6, x = 7, y-2=0, etc. which are of first degree in variables x and both or only in x or only in y are known as linear equations and each equation always represents a straight line.

Inclination of a straight line:

The angle that a straight line makes with the positive direction of the x-axis, measured in the anticlockwise directions is called the inclination of the line. The inclination is usually denoted by θ.

Slope (or gradient) of a straight line:

If θ is the inclination of a line, then tan θ is called its slope (or gradient) of the line.

The slope of the x-axis or a line parallel to the X-axis is zero.

The slope of the y-axis or a line parallel to the y-axis is tan 90° = ∞ i.e., not defined.

The slope of a line is positive if it makes an acute angle in the anti-clockwise direction with the x-axis.

The slope of a line is negative if it makes an obtuse angle in the anticlockwise direction with the x-axis.

Equation of a line:

Slope- Intercept Form: y= mx+c

Point-Slope Form: $y-y_{1}= m(x-x_{1})$

Two-Points Form; $y-y_{1}= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$

Equation of a straight line parallel to x-axis: y = b

Equation of a straight line parallel to y-axis: x= a

Solutions of MCQs are given below. I suggest you give the test first in Google Form and then go through the solution. That way you will understand better what you have done wrong.

Click here to open MCQs of Straight Lines in Google Form

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MCQ Class 10 Mathematics

Banking (Set 1)

Banking (Set 2)

GST

Linear Inequations (Set 1)

Linear Inequations (Set 2)

Quadratic Equations

Ratio and Proportion

Matrices

Arithmetic Progression (A.P.)

Mock Question Paper for Semester 1 (2021) ICSE Class X Mathematics

ICSE Class X – Specimen Question Paper released by CISCE for Semester 1 (2021) – Mathematics

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Solutions :

Question 1: The slope of the line passing through the points (0, −3) and (2,1) is

(a) 0 (b) 1/2 (c) 2 (d) not defined

Solution : $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1+3}{2-0}=2$

option (c)

Question 2: The inclination of the line ${\color{Blue} x-\sqrt{3}\, \, y+2\, \sqrt{3}\, =0}$ is

(a) 0° (b) 30° (c) 60° (d) 90°

Solution: $x-\sqrt{3}\, \, y\, +2\, \sqrt{3}=0\: \Rightarrow \sqrt{3}\, \, y=x+2\, \sqrt{3}$

$y=\frac{1}{\sqrt{3}}\: x+2\: \therefore \, m=\frac{1}{\sqrt{3}} \: \: and\: \: \theta =30^{0}$

option (b)

Question: The slope of a line parallel to line ${\color{Blue} 4x-5y-9=0}$ is

(a) 0 (b) 5/4 (c) 4/5 (d) not defined

Solution: $4x-5y-9=0\: \: \Rightarrow \: 5y=4x-9\: \Rightarrow \, y=\frac{4}{5}x-9$

$m=\frac{4}{5}$

option (c)

Question 4: The equation of a line passing through the point (−2, 3) and having x-intercept 4 units is

(a) x+2y = 8 (b) x+2y=4 (c) 2x+y=8 (d) x+y+4=0

Solution: $y-3=\frac{0-3}{4+2}(x+2)\: \: \:\: \: using \: \: y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$

$\Rightarrow y-3=-\frac{1}{2}(x+2)\: \: \Rightarrow 2y-6=-x-2\: \: \Rightarrow x+2y=4$

option (b)

Question 5: If the lines 2x + 3y − 7 = 0 and 4y − px − 12 = 0 are perpendicular to each other , then the value of p is

(a) 6 (b) − (8⁄3) (c) − 6 (d) 3⁄8

Solution : $2x+3y-7=0\: \: \Rightarrow \: y=-\frac{2}{3}x+\frac{7}{3}\: \:\: \Rightarrow m_{1}=-\frac{2}{3}$

$4y-px-12=0\: \:\: \Rightarrow y=\frac{p}{4}x+3\: \: \Rightarrow m_{2}=\frac{p}{4}$

As two lines are perpendicular, $m_{1}\times m_{2}=-1\: \: \Rightarrow -\frac{2}{3}\times \frac{p}{4}= -1\: \: \Rightarrow p=-6$

option (c)

Question 6: The equation of the line parallel to the line 3x+2y=8 and passing through the point (0,1) is

(a) 2x − 3y − 3= 0 (b) 3x + 2y − 2 = 0 (c) 3x + 2y + 2 = 0 (d) 2x − 3y + 3 = 0

Solution: $3x + 2y=8\: \: \Rightarrow y=-\frac{3}{2}+\frac{8}{3}\: \: \Rightarrow m=-\frac{3}{2}$

So the slope of the required equation = $-\frac{3}{2}$ as it is parallel to the given equation and it passes through (0,1).

The required equation is $y-1=-\frac{3}{2}(x-0)\: \: \: \: using\: \: y-y_{1}= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$

$\Rightarrow \: 2y-2=-3x\: \: \: \Rightarrow 3x+2y-2=0$

option (b)

Question 7: The equation of the line AB parallel to the x-axis as shown in the figure is

(a) x = 2 (b) y = 2 (c) x = – 2 (d) y = – 2

Solution : Straight line is parallel to x-axis.

It’s equation is of the form $y=k$

From the above figure, the required equation is $y = 2$

option (b)

Question 8: The equation of the line CD is

(a) 5x + 3y =15 (b) 3x − 5y = 15 (c) 3x +5y = 15 (d) 5x − 3y = 15

Solution: From above figure, CD passes through (5,0) and (0,3).

So the equation of line CD is $y-0=\frac{0-3}{5-0}(x-5)\: \: \: \: using \: \: y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$

$\Rightarrow \: \: 5y=-3x+15\: \: \: \Rightarrow 3x+5y=15$

option (c)

Question 9: The equation of the line PQ is

(a) $y=x+3$ (b) $y=\sqrt{3}x-3$ (c) $y=\frac{1}{\sqrt{3}}\, x-3$ (d) $y=x-3$

Solution: From above figure, m = tan 45° = 1 , c = − 3

Equation of PQ is y = x − 3 using y = mx + c

option (d)

Question 10: The equation of a line is 3x + 4y − 7 = 0.

(i) The slope of the line is

(a) − (3⁄4) (b) 4⁄3 (c) −(4⁄3) (d) 3⁄4

(ii) The equation of a given line perpendicular to the given line and passing through the intersection of the

lines x − y + 2 = 0 and 3x + y − 10 = 0 is

(a) 3x + 4y = 22 (b) 4x − 3y + 4 = 0 (c) 4x + 3y = 20 (d) 3x − 4y + 10 = 0

Solution : (i) $3x+4y-7=0\: \: \: \Rightarrow 4y=-3x+7\: \: \Rightarrow y=-\frac{3}{4}\: x+\frac{7}{4}$ ∴ $m= -\frac{3}{4}$

option (a)

(ii) Solving these two equations, x – y+2 = o and 3x + y – 10 = 0 , we get x = 2, y = 4.

So required straight line passes through (2, 4).

Slope of the line 3x + 4y – 7 = 0 is m = − (3⁄4). As the required line is perpendicular to this line, the slope of the required straight line is 3⁄4.

∴ The required equation is $y-4=\frac{4}{3}(x-2)\: \: using \: \: \: y-y_{1}=m(x-x_{1})$

$\Rightarrow \: 3y-12=4x-8\: \: \: \Rightarrow 4x-3y+4=0$

option (b)

Question 11: ${\color{Blue} L_{1},\: L_{2},\: L_{3}\: and\: L_{4}}$ are 4 straight lines as shown in the figure.

(i) The equation of the straight line ${\color{Blue} L_{1}}$ is

(a) y = 2x (b) y − 2x + 2 = 0 (c) 3x + 2y = 6 (d) x = − 5

(ii) The equation of the straight line ${\color{Blue} L_{2}}$ is

(a) y = 2x (b) y − 2x + 2 = 0 (c) 3x + 2y = 6 (d) x = − 5

(iii) The equation of the straight line ${\color{Blue} L_{3}}$ is

(a) y = 2x (b) y − 2x + 2 = 0 (c) 3x + 2y = 6 (d) x = − 5

(iv) The equation of the straight line ${\color{Blue} L_{4}}$ is

(a) y = 2x (b) y − 2x + 2 = 0 (c) 3x + 2y = 6 (d) x = − 5

Solution: (i) $L_{1}$ is parallel to y-axis and it it is in the left side of y-axis. Its equation is of the form x = k. So its equation is x = − 5.

option (d)

(ii) $L_{2}$ is making an obtuse angle with x-axis, so its slope is negative and its y-intercept is positive.

$3x+2y=6\: \: \Rightarrow y=-\frac{3}{2}\: x+3\: \:\: \therefore m=-\frac{3}{2}\: \: and\: \: c=+3$

Clearly $3x+2y=6$ is the equation of $L_{2}.$

option (c)

(iii) $L_{3}$ is passing through the origin. Its y-intercept is zero, i.e., c = 0. It’s equation is of the form y = mx. So it’s equation is y = 2x.

option (a)

(iv) $L_{4}$ is making an acute angle with x-axis. So its slope m is positive. It’s y-intercept (c) is negative.

$y-2x+2=0\: \: \Rightarrow y=2x-2\: \: \therefore \: m=2\: and\: c=-2$ . Clearly $y-2x+2=0$ is the equation of $L_{4}.$

option (b)

Question 12: The equation of the line whose gradient is ${\color{Blue} \frac{3}{2}}$ and which passes through P, where P divides the line segment joining A(−2, 6) and B(3, −4) in the ratio 2 : 3 is

(a) 3x − 2y = 3 (b) 3x − 2y + 4 = 0 (c) 2x + 3y = 6 (d) 3x – 2y = 6

Solution:

$P= (\frac{2\times 3+3\times (-2)}{2+3},\: \frac{2\times (-4)+3\times 6}{2+3})=(0, \, 2)$

So the equation of the line whose slope is 3⁄2 and passing through (0, 2) is

$y-2=\frac{3}{2}(x-0)\: \: \: using \: \: y-y_{1}= m(x-x_{1})$

$\Rightarrow \: 2y-4=3x\: \: \Rightarrow \: 3x-2y+4=0$

option (b)

Question 13: Answer the following questions from the given figure:

(i) The coordinates of A

(a) (2, 3) (b) (3, 2) (c) (3, 0) (d) (3, 3)

(ii) The coordinates of B

(a) (−1, 0) (b) (−1, 2) (c) (1, 2) (d) (3, 0)

(iii) The coordinates of C

(a) (0, 3) (b) (2, 3) (c) (3, 0) (d) (−1, 0)

(iv) The slope of BC

(a) 3 (b) −2 (c) 2 (d) − ½

(v) The equation of the line through A and parallel to BC.

(a) 2x − y = 1 (b) x + 2y = 10 (c) x + 2y = 8 (d) x − 2y + 4 = 0

Solution: (i) A = (2, 3)

option (a)

(ii) B = (−1, 2)

option (b)

(iii) C = (3, 0)

option (c)

(iv) Slope of BC = $\frac{0-2}{3+1}=-\frac{1}{2}$

option (d)

(v) $m=-\frac{1}{2}\: and\: \: (x_{1},y_{1})=(2,3)$

The equation of the line through A and parallel to BC is $y-3=-\frac{1}{2}(x-2)\: \: \Rightarrow x+2y=8$

option (c)

Question 14: The given figure (not drawn to scale) shows two straight lines AB and CD. The equation of line AB is y = x + 1 and equation of line CD is y = √3 x − 1.