Find an equation of the straight line that is perpendicular to the straight line x + 2y = 5 and that passes through the point (3, 7).
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Answered by
2
Step-by-step explanation:
refer to the attachment for answer.
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Answered by
11
Answer:
Correct option is
D
4x−3y+2=0
The line x+2y=5 and 3x+7y=17 intersect at the point (1,2)
Slope of line 3x+4y=10 is m=−
4
3
The slope of line perpendicular to 3x+4y=10 is m
′
=
m
−1
=
⎝
⎜
⎜
⎛
−
4
3
−1
⎠
⎟
⎟
⎞
=
3
4
So, the equation of line through (1,2) and having slope
3
4
is
(y−2)=
3
4
(x−1)
⟹3(y−2)=4(x−1)
⟹4x−3y+2=0
The answer is option (D)
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