Question

the line 2y=x+2 meets the curve 3x²+xy-y²=12 at the points A and B find the coordinates of the points a and b
given that the point c has coordinates (0,6), show that the triangle ABC IS right angled
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Answer 1

Answer:

A( - 2, 0) and B(2 , 2)

Step-by-step explanation:

To find the coordinates of points A($x_{1}$ , $y_{1}$) and B($x_{2}$ , $y_{2}$) we have to solve the system of equations

2y = x + 2 .... (1)

3x² + xy - y² = 12 ..... (2)

From (1) → (x = 2y - 2) → (2)

3(2y - 2)² + (2y - 2)y - (2y - 2)² = 12

3(4y² - 8y + 4) + 2y² - 2y - y² = 12

12y² - 24y + 12 + 2y² - 2y - y² - 12 = 0

13y² - 26y = 0

13y(y - 2) = 0

$y_{1}$ = 0 ⇒ 0 = x + 2 , $x_{1}$ = - 2

$y_{2}$ = 2 ⇒ 4 = x + 2 , $x_{2}$ = 2

A( - 2, 0) and B(2 , 2)

Answer 2

To find coordinates of points of intersection of the given line and curve, we've to solve for x and y for their equations.

From equation of the line we get,

$\longrightarrow y=\dfrac{x+2}{2}$

$\longrightarrow y=\dfrac{x}{2}+1$

Putting this value of y in the equation of the curve,

$\longrightarrow 3x^2+x\left(\dfrac{x}{2}+1\right)-\left(\dfrac{x}{2}+1\right)^2-12=0$

$\longrightarrow 3x^2+\dfrac{x^2}{2}+x-\dfrac{x^2}{4}-x-1-12=0$

$\longrightarrow\dfrac{13x^2}{4}-13=0$

$\longrightarrow x^2-4=0$

$\longrightarrow x=\pm 2$

For $x=2,$

$\longrightarrow y=\dfrac{2}{2}+1$

$\longrightarrow y=2$

For $x=-2,$

$\longrightarrow y=\dfrac{-2}{2}+1$

$\longrightarrow y=0$

Hence the coordinates of the points A and B are (2, 2) and (-2, 0).

Now we have three points A(2, 2), B(-2, 0) and C(0, 6).

Slope of line joining A and B is,

$\longrightarrow m_{AB}=\dfrac{2-0}{2-(-2)}$

$\longrightarrow m_{AB}=\dfrac{1}{2}$

Slope of line joining A and C is,

$\longrightarrow m_{AC}=\dfrac{2-6}{2-0}$

$\longrightarrow m_{AC}=-2$

We see the product of slopes of lines AB and AC is -1.

$\longrightarrow \dfrac{1}{2}\times-2=-1$

This means the lines AB and AC are perpendicular to each other.

$\begin{minipage}{105mm}``\textit{Two lines are perpendicular to each other if and only if the product of their slopes equals -1 .}"\end{minipage}$

This implies the triangle ABC is a right triangle, right angled at A.

Hence Proved!