Question

NCERT Solutions Class 11 Maths Chapter 10 Straight Lines is one of the important chapters of mathematics. We have provided solutions for the entire chapter in in-depth detail.

These NCERT Solutions Class 11 is prepared by the subject matter experts at Sarthaks.

• Straight Lines – a straight line is an infinite one-dimensional figure which has no width. Straight-line has no curves in it. A straight line can be vertical, horizontal, or slanted.

• Two-dimensional geometries – shape or figure on a two-dimensional plane that have only length and width. Two-dimensional figures have two faces. Example: circle, triangle, square, rectangle, pentagon.

• Shifting of origin – during shifting of origin we choose a new point on the coordinate plane which acts as a new origin of the plane. Now new distance is measured from the new origin.

• Slope of a line and angle between two lines – the angle between two lines helps us to quantify the relationship between two lines. The angle between two lines is the measure of inclination of two lines with each other.

• Various forms of equations of a line –

Slope (m) of a non-vertical line passing through the points (x1 , y1 ) and (x2, y2 )	m=(y2-y1)/(x2-x1), x1≠x2
Equation of a horizontal line	y = a or y=-a
Equation of a vertical line	x=b or x=-b
Equation of the line passing through the points (x1 , y1 ) and (x2, y2 )	y-y1= [(y2-y1)/(x2-x1)]×(x-x1)
Equation of line with slope m and intercept c	y = mx+c
Equation of line with slope m makes x-intercept d.	y = m (x – d).
Intercept the form of the equation of a line	(x/a)+(y/b)=1
The normal form of the equation of a line	x cos α+y sin α = p

• Parallel to axis – a straight line that makes no angle with the axis such lines are parallel to the axis. The distance of the line from the axis is always constant.

• Slope-intercept form - in slope intercept form the equation of line is y = mx + c.

• General equation of a line – for the first-degree equation of two variables the general equation of a line is represented as Ax + By + C = 0, where A, B is not equal to 0 and A, B and C are constant which belongs to real numbers.

Our NCERT Solutions Class 11 Maths has a unique approach to the solutions in an organized manner.

Answer 1

74. Prove that the product of the lengths of the perpendiculars drawn from the points \((\sqrt{a^2-b^2},0)\) and \((-\sqrt{a^2-b^2},0)\) to the line \(\frac xacos\theta +\frac yb sin \theta=1 \) is b².

Answer:

The equation of the given line is

Length of the perpendicular from point \((\sqrt{a^2-b^2},0)\) to line (1) is

Length of the perpendicular from point \((-\sqrt{a^2-b^2},0)\) to line (2) is

On multiplying equations (2) and (3), we obtain

Hence, proved.

75. A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

Answer:

The equations of the given lines are

2x – 3y + 4 = 0 ………………. (1)

3x + 4y – 5 = 0 ………………. (2)

6x – 7y + 8 = 0 ……………… (3)

The person is standing at the junction of the paths represented by lines (1) and (2).

On solving equations (1) and (2), we obtain 

Thus, the person is standing at point \(\left(-\frac{1}{17},\frac{22}{17}\right)\)

The person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point \(\left(-\frac{1}{17},\frac{22}{17}\right)\)

Slope of the line (3) = \(\frac67\)

∴ Slope of the line perpendicular to line (3) = \(\cfrac1{(\frac67)}=-\frac76\)

The equation of the line passing through \(\left(-\frac{1}{17},\frac{22}{17}\right)\) and having a slope of \(-\frac76\) is given by 

Hence, the path that the person should follow is 119x + 102y = 125.