Question

Find the equation of the altitude of a triangle XYZ drawn from the vertex Y upon XZ whose vertices are X(3,2) ,Y(-2,1) , Z(2,-1)
(Ans is : 3x+y+4=0) how ?

Answer 1

Answer:

Solution:-

Let ABC be the triangle with vertices A(2,−2),B(1,1) and C(−1,0)&AD be the altitude of △ABC drawn from A.

Let m

1

&m

2

be the slope of line AD and BC respectively.

Now, AD⊥BC

∴m

1

×m

2

=−1

⇒m

1

=

m

2

−1

⟶(i)

Slope of line BC-

Slope of a line joining points (x

1

,y

1

)&(x

2

,y

2

)=

x

2

−x

1

y

2

−y

1

∴ Slope of BC joining B(1,1)&C(−1,0)=

−1−1

0−1

=

−2

−1

=

2

1

On substituting the value of m

2

in eq

n

(i), we get

m

1

=

(

2

1

)

−1

=−2

The equation of line passing through the point (x

1

,y

1

) with slope m is-

y−y

1

=m(x−x

1

)

∴ Equation of altitude AD passing through A(2,−2) with slope 2 is-

y−(−2)=−2×(x−2)

⇒y+2=−2x+4

⇒y=−2x+2

Answer 2

Answer:

thik a 3

Step-by-step explanation:

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