Question

256. The equation y − 5 = m(x − 2) represents a line, no matter what value m has. (a) What are the x- and y-intercepts of this line? (b) For what value of m does this line form a triangle of area 36 with the positive axes? (c) Show that the area of a first-quadrant triangle formed by this line must be at least 20.

Answer 1

(a)

The given equation is y − 5 = m(x − 2)

To find the Y-intercepts

Put 0 for X variable in the equation

ie, Y-5 = -2m

Y = -2m+5

The Y-intercepts is (0, -2m+5)

Similarly put 0 for Y in the equation to find the X-intercepts

-5 = mX -2m

mX = 2m-5

X = (2m-5)/m

The X-intercepts is ((2m-5)/m, 0)

(b)

The value of m does this line form a triangle of area 36 with the positive axes is,

m = -1/2 , - 25/2

ie, The triangle formed by the X and Y axis with the line has to achieve the area of 36

AB is the given line

The area of the triangle ABC, A = 1 /2 bh = 36 (Given)

1/2 X (-2m+5) X (2m-5/m) = 36

solving the values of m, we get m = -1/2 , - 25/2

(c)

Condition is the area should be greater than or equal to 20

A = 1 /2 bh = 20

1/2 X (-2m+5) X (2m-5/m) = 20

ie,

solving the values of m, we get m = -5/2

we can see that any other values of m, which has the area larger than 20