Question

if the coordinates of the vertex A of a square ABCD are (3,2)of the equation of the digonalBD is 3x-7y+6=0 find the equation of the diogonal ac ,also,find the coordinates of the centre of the square

Answer 1

Answer:

Here is your answer

Step-by-step explanation:

$a \: =$ (3,2)

BD = $3x - 7y + 6 = 0$

So, we need to find the slope of BD.

$7y = 3x + 6$

$y = \frac{3x + 6}{7}$

$y = \frac{3}{7} x + \frac{6}{7}$

So, the slope(m) is = $\frac{3}{7}$

Now, the slope(n) of a perpendicular line can be found by

$m \times n = - 1$

$\frac{3}{7} \times n = - 1$

$n = - \frac{7}{3}$

Thus, the family equation of AC is

$y = - \frac{7}{3} x + c$

Putting A(3,2) in the above equation,

$2 = - \frac{7}{3} \times 3 + c$

$c = 2 + 7$

$c = 9$

So, AC may be written as,

$y = - \frac{7}{3} x + 9$

To find the central point, we need to find the intersection point of the two lines BD and AC.

$\frac{3x + 6}{7} = - \frac{7}{3}x + 9$

_-_-_-_-

$x( \frac{3}{7} + \frac{7}{3} ) = 9 - \frac{6}{7}$

Solve for $x$ and $y$.

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