the angle of a triangle ABC are towards - 5 x minus 3 and X find the value of x and hence find the angles of triangle ABC
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The given lines are as follows:
x+y=6
5x−3y+2=0
x+y−6
The given sides AB,BC and CA of a triangle ABC are :
5x−3y+2=0−−−−−−−−(1)
x−3y−2=0−−−−−−−−−−−(2)
x+y−6=0−−−−−−−−−−−−(3)
on solving (1) and (2), we get, x=2,y=4
Thus, AB and CA intersect at A(2,4)
Let AD be the altitude.
Hence, AD⊥BC
∴ slope of AD× slope of BC=−1
Here, slope of BC= 1/3
∴ slope of AD=−3
Hence, the equation of the altitude AD passing through A(2,4) and having slope −3 is y−4=−3(x−2) ⇒3x+y=10.
Hence, equation of the altitude through the center A is 3x+y=10
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