Question

"find the equation of the straight line passing through the point (-6,2) and perpendicular to the line joining the points (-3,2) and (5,-3) ."

can any one solve it it's from class 10 coordinate geometry chapter please solve it step by step

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Answer 1

Correct option is

B

x + y - 5 = 0

The equation of any straight line can be written as y=mx+c, where m

is its slope and c is its y - intercept.

Slope of the line passing through points (x

1

,y

1

) and (x

2

,y

2

) =

x

2

−x

1

y

2

−y

1

So, slope of the line joining (−5,6),(−6,5)=

5−6

−6+5

=1

Slope of the line perpendicular to this line =−1

So, equation of the required line will be y=−x+c

Since, this line passes through (2,3), on substituting (2,3) in the equation we get

3=−2+c

=>c=5

Hence, required equation of the line is y=−x+5 or x+y−5=0