The ends of the base of an isosceles triangle

Asked by admin @ 17/08/2021 in Math viewed by 244 People

The ends of the base of the isosceles triangle are at (2a,0)and (0,a). the equation of one side is x = 2a . the equation of the other side is

1.x + 2y –a =0
2. x + 2y –2a =0
3.3x + 4y -4a =0
4. 3x + 4y +4a =0

Answered by admin @ 17/08/2021

Answer:

Equation of other side: 3x-4y+4a=0

Step-by-step explanation:

Please take a look with attachment for triangle.

The ends of the base of the isosceles triangle are at B (2a,0)and C (0,a).

Mid point of B(2a,0) and C(0,a) is M

$\text{Mid point M: }(a,\frac{a}{2})$

$\text{Slope of base BC}=\frac{a-0}{0-2a}\Rightarrow -\frac{1}{2}$

Now we find the equation of perpendicular bisector equation of BC which passes through the point M.

Slope of perpendicular line = 2 and passing point M(a,a/2)

Equation of line using point slope form.

$y-\frac{a}{2}=2(x-a)$

$y=2x-\frac{3a}{2}$

we are given the equation of one side is x=2a

If we find the intersection point of x=2a and $y=2x-\frac{3a}{2}$

We will get the third vertices of isosceles triangle

$y=2(2a)-\frac{3a}{2}\Rightarrow \frac{5a}{2}$

$\text{Third vertices of triangle A:} (2a,\frac{5a}{2})$

Now we find equation of line AC where A(2a,5a/2) and C(0,a)

Equation of line AC using two point formula

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

$\frac{y-a}{x-0}=\frac{5a/2-a}{2a-0}$

$y-a=x(\frac{3}{4})\Rightarrow y=\frac{3}{4}x+a$

Now we write equation in standard form:

3x-4y+4a=0

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