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CLASS_11_MATHS_SOLUTIONS_NCERT

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Class XI Chapter 10 – Straight Lines Maths<br />

______________________________________________________________________________<br />

On solving equation (1) and (2), we obtain<br />

2m<br />

5m2<br />

x and y <br />

m1 m1<br />

2m<br />

5m2<br />

<br />

,<br />

m1 m1<br />

is the point of intersection of line (1) and (2).<br />

<br />

Since this point is at a distance of 3 units from points (-1, 2), accordingly to distance formula,<br />

m1 m1<br />

<br />

<br />

2 2<br />

2m<br />

5m2<br />

<br />

1 2 3<br />

m1 <br />

m1<br />

<br />

<br />

2 2<br />

2<br />

2 2<br />

2m m1 5m 22m<br />

2 <br />

3<br />

m 1<br />

m 1<br />

<br />

<br />

2<br />

9 9m<br />

9<br />

2<br />

1<br />

m<br />

<br />

m1<br />

1<br />

2<br />

2 2<br />

1 m m 1<br />

2m<br />

2m<br />

0<br />

m<br />

0<br />

Thus, the slope of the required line must be zero i.e., the line must be parallel to the x-axis.<br />

Question 18:<br />

Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a<br />

plane mirror.<br />

Solution 18:<br />

The equation of the given line is<br />

x + 3y = 7 …(1)<br />

Let point B (a, b) be the image of point A (3, 8).<br />

Accordingly, line (1) is the perpendicular bisector of AB.<br />

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