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Published on: 16/09/2019
Straight Lines
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Questions + Answers key
Take MCQ Mathematics Test
1.
Find the perpendicular distance of the point of intersection of the lines 2x + 3y - 7 = 0, 3x + 4y - 10 = 0 from the line 2x - 4y + 10 = 0.
2.
Find the equation of a straight line perpendicular to the line joining the points (1, -3) and (2, 1) and cutting off intercept - 4 from the x-axis.
3.
Find the new coordinates of a point (-3, 4) when the origin is shifted to(-1,2).
4.
Find the transformed equal to x 2+2x - 2y2+2 = 0 when the orgin is shifted to (-3, 2).
5.
Find the slope of the line which makes an angle of 60° with the line 2x - y + 7 = 0.
6.
If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.
7.
Find the equations of the medians of a triangle, the coordinates of whose vertices are (-1, 6), (-3, -9) and (5, -8).
8.
In what ratio does the line joining the points (2, 3) and (4, 1) divide the segment joining the points (1, 2) and (4, 3)?
9.
Find the equation of a straight line which cuts off an intercept of 3 units on negative direction of y-axis and makes an angle of 120° with the positive direction of x-axis.
10.
What is the value of x so that the line through (3, x) and (2, 7) is parallel to the line through (-1, 4) and (0, 6).
11.
If the angle between two lines is \(\pi\over 4\) and slope of one of the lines is 2. Find slope of another line.
12.
Find the values of \(\theta \) and p, if the equation xcos \(\theta \) +ysin \(\theta \) =p is the normal form of the line \(\sqrt { 3x } \)+y+2=0
13.
Find the equation of the straight line which bisects the distance between the points A(a, b), B(a', b') and also bisects the distance between the point C (- a, b) and D(a', - b').
14.
A line passes through the points A(4,-6) and B(-2,-5). Show that the line AB makes an obtuse angle with the X-axis.
15.
Reduce the following equation into slope intercept form and find their slopes and the y-intercepts y=0.
1.
√5
2.
x + 4y + 4 = 0
3.
(-2, 2)
4.
x2+2xy-2y2-2x-14y+15=0
5.
\(\frac{-8\pm5\sqrt{3}}{11}\)
6.
\(m=\frac{1\pm 5\sqrt{2}}{7}\)
7.
29x + 4y + 5 = 0, 8x - 5y - 21 = 0, 13x + 14y + 47 = 0.
8.
1: 1
9.
\(\sqrt{3}x+y+3=0\)
10.
x = 9
11.
\(-3,\frac{1}{3}\)
12.
Given equation of line is \(\sqrt { 3x } \) + y+ 2 =0
\(\text { This equation can be reduced as }\) -\(\sqrt { 3x }\) - y=2
on dividing both sides of Eq (i) b
\(\sqrt { (-\sqrt { 3 } )^{ 2 }+(-1)^{ 2 } } =2\) we get
\(-\frac{\sqrt{3}}{2} x-\frac{1}{2} y=\frac{2}{2} \)
\(\Rightarrow\left(-\frac{\sqrt{3}}{2}\right) x+\left(-\frac{1}{2}\right) y=1\)
\(\text { On comparing equation (1) to } x \cos \theta+y \sin \theta=p \text { . we obtain }\)
\(\cos \theta=-\frac{\sqrt{3}}{2}, \sin \theta=-\frac{1}{2}, \text { and } p=1\)
\(\text { Since the values of } \sin \theta \text { and } \cos \theta \text { are negative, } \theta=\pi+\frac{\pi}{6}=\frac{7 \pi}{6}\)
\(\text { Thus, the respective values of } \theta \text { and } p \text { are } \frac{7 \pi}{6} \text { and } 1\)
13.
Let the line MN bisects the line AB at M and CD at N.
Coordinates of M = \(\left( \frac { a+{ a }^{ ' } }{ 2 } ,\frac { b+{ b }^{ ' } }{ 2 } \right) \)
and coordinates of N = \(\left( \frac { -a+{ a }^{ ' } }{ 2 } ,\frac { b-{ b }^{ ' } }{ 2 } \right) \)
Now, equation of line MN is
\(\left( y-\frac { b+{ b }^{ ' } }{ 2 } \right) =\left( \frac { \frac { b-{ b }^{ ' } }{ 2 } -\frac { b+{ b }^{ ' } }{ 2 } }{ \frac { -a+{ a }^{ ' } }{ 2 } -\frac { a+{ a }^{ ' } }{ 2 } } \right) \left( x-\frac { a+{ a }^{ ' } }{ 2 } \right) \)
Ans. 2b'x - 2ay + ab - a'b' = 0
14.
Slope of AB is negative.
15.
The given equation is y = 0.
It can be written as
y = 0.x + 0
This equation is of the form y = mx + c, where m = 0 and c = 0.
Therefore, equation (3) is in the slope-intercept form, where the slope and the y-intercept are 0 and 0 respectively.
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