SHOW THAT THE TRIANGLE WHOSE VERTICES ARE A 4 5 B 3 9 AND C

SHOW THAT THE TRIANGLE WHOSE VERTICES ARE A (4, 5), B (- 3, 9), AND C (1, 3) IS A RIGHT TRIANGLE. FIND ITS AREA.

Solution:

Slope of AB = y₂-y₁ / x₂- x₁

=> slope of AB = 4/-7 => -4/7

and

Slope of BC = y₂-y₁ / x₂- x₁

=> slope of BC = -6/4 => -3/2

similarly;

Slope of AC = y₂-y₁ / x₂- x₁

=> slope of AC = -2/-3 => 2/3

here :
slope of AC× slope of BC = -1

that means AC and BC are prependicular to each other.

i.e, right angle at C.

Area of ABC = (1/2)×length of side AC ×length of side BC

=> Area of ABC = (1/2) × 13 × 52

=> Area = 13 sq-units