CLASS_11_MATHS_SOLUTIONS_NCERT

Class XI Chapter 10 – Straight Lines Maths
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The equation of the line passing through point (2,3) and having a slope of
2
3
y-3 = x-2
2 3 1
y x
3 2 x2 3 1
y 4 2 3 6 3 3
3 2 x2 3 1
y 1
8 3
2 3 1 3 2 3 1 2 3 2 3 2
2
3
2 3 1
is
In this case, the equation of the other line is
3 2
x + 2 3 1
y = -1 + 8 3
2
3
Case II : m2
2 3 1
2
3
The equation of the line passing through points 2,3 and having a slope of is
2 3 1
2
3
y-3 = x-2
2 3 1
2 3 1 y32 3 1 2 3 1 x 22 3 1
2 3 1 y2 3 1
x 4 2 3 6 3 3
2 3 x2 3 1
y18 3
If this case, the equation of the other line is 2 3 x2 3 1
y18 3
Thus, the required equation of the other line is 3 2 x 2 3 1 y 1 8 3or 2 3 x 2 3 1 y 1
8 3
Question 13:
Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).
Solution 13:
The right bisector of a line segment bisects the line segment at 90 .
The end-points of the line segment are given as A (3, 4) and B (-1, 2).
31 42 Accordingly, mid-point of AB = , 1,3
2 0
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