Question

14. Find the length of perpendiculars drawn from the origin on the sides of the triangle whose
vertices are A (2,1), B (3,2) and C (-1,-1).
2​

Answer 1

Step-by-step explanation:

ER

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