NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3.
Reduce the following equations into slope- intercept form and find their slopes and the y-intercepts.
Hence the slope is and the y-intercept = 0.
m = 0, c = 0.
Hence, slope is 0 and the y-intercept is 0.
Reduce the following equations into intercept form and find their intercepts on the axes.
On comparing (1) and (2), we get a = 4, b = 6 Hence, the intercepts of the line are 4 and 6.
Hence, the intercepts of the line are and -2.
The above equation shows that, it is not the required equation of the intercept form as it is parallel to x-axis.
We observe that y-intercept of the line is , but there is no intercept on x-axis.
Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.
⇒ – cos 60°x + sin 60° y = 4.
⍵ Distance of the line from origin is 4 and the angle between perpendicular and positive x-axis is 120°.
⍵ Distance of the line from origin is 2 and the angle between perpendicular and positive x-axis is 90°.
is the required equation in normal form.
∴ Distance of the line from the origin is and the angle between perpendicular and the positive x-axis is 315°.
Find the distance of the point (-1, 1) from the line 12 (x+ 6) = 5(y — 2).
Find the points on the x-axis, whose distances from the line are 4 units.
Let (a, 0) be the point on x-axis whose distance from line (i) is 4 units.
(ii) |(x + y) + p = 0 and |(x + y) – r = 0.
Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (-2, 3).
Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.
⇒ 7x + y – 21 = 0, is the required equation of line.
Find angles between the lines + y = 1 and x + = 1.
Since we have to find an angle between the two lines i.e., firstly we have to find the slopes of (i) and (ii).
The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle. Find the value of h.
Given points are (h, 3) and (4,1).
Prove that the line through the point (x1 y1) and parallel to the line Ax + By + C = 0 is A(x-x1) + B(y-y1) = 0.
=> A(x – x1) + B(y – y1) = 0.
Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.
We have given a point (2, 3), through which two lines are passing and intersects at an angle of 60°.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).
suppose the given points are A and B.
Let M be the mid point of AB.
Find the coordinates of the foot of perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.
The perpendicular from the origin to the line y = mx + c meets it at the point (-1,2). Find the values of m and c.
If p and q are the lengths of perpendiculars from the origin to the lines x cosθ – y sinθ = k cos 2θ and x secθ + y cosecθ = k, respectively, prove that p2 + 4q2 = k2.
Given p and q are the lengths of perpendiculars from the origin to the lines x cos θ – ysinθ=k cos 2θ and xsecθ+y cosec θ = k.
In the triangle ABC with vertices A(2, 3), 8(4, -1) and C( 1, 2), find the equation and length of altitude from the vertex A.
If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that .
We hope the NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3, help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3, drop a comment below and we will get back to you at the earliest.