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The equation of the diagonal AC of given square ABCD is 3x-4y+10=0
and the coordinates of vertex B are (4,-5). find the equation of diagonal BD
Answer
4x+3y-1=0​

The equation of the diagonal AC of given square ABCD is 3x4y100and the coordinates of vertex B are 45 find the equation of diagonal BD Answer 4x3y10 class=

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Answer:

see explanation

Step-by-step explanation:

In a square

• diagonals are perpendicular bisectors of each other

Thus diagonals AC and BD are perpendicular

Given the equation of a line in the general form

Ax + By + C = 0 , then

slope m = [tex]\frac{-A}{B}[/tex]

Given equation of diagonal AC

3x - 4y + 10 = 0 ← in general form

with A = 3 and B = - 4 , then

slope m = [tex]\frac{-3}{-4}[/tex] =  [tex]\frac{3}{4}[/tex]

given a line with slope m, then the slope of a line perpendicular to it is

[tex]m_{perpendicular}[/tex] = - [tex]\frac{1}{m}[/tex] = - [tex]\frac{1}{\frac{3}{4} }[/tex] = - [tex]\frac{4}{3}[/tex]

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

the partial equation of diagonal BD is then

y = - [tex]\frac{4}{3}[/tex] x + c

to find c, substitute B (4, - 5 ) for x and y into the partial equation

- 5 = - [tex]\frac{4}{3}[/tex] (4) + c = - [tex]\frac{16}{3}[/tex] + c ( add [tex]\frac{16}{3}[/tex] to both sides )

- [tex]\frac{15}{3}[/tex] + [tex]\frac{16}{3}[/tex] = c ⇒ c = [tex]\frac{1}{3}[/tex]

y = - [tex]\frac{4}{3}[/tex] x + [tex]\frac{1}{3}[/tex] ← equation of diagonal BD

multiply through by 3 to clear the fractions )

3y = - 4x + 1 ( add 4x to both sides )

4x + 3y = 1 ( subtract 1 from both sides )

4x + 3y - 1 = 0 ← equation of diagonal BD

semsee45

Answer:

[tex]4x+3y-1=0[/tex]

Step-by-step explanation:

In a square, the diagonals are perpendicular bisectors of each other.

Given that the equation of diagonal AC is 3x - 4y + 10 = 0, we can determine the slope of AC and then find its negative reciprocal to get the slope of diagonal BD.

Rearrange the given equation 3x - 4y + 10 = 0 into slope-intercept form, y = mx + b, where m is the slope:

[tex]\begin{aligned}3x - 4y + 10 &= 0\\\\4y &= 3x + 10\\\\y &= \dfrac{3}{4}x + \dfrac{10}{4}\\\\y &= \dfrac{3}{4}x + \dfrac{5}{2}\end{aligned}[/tex]

So, the slope of diagonal AC is:

[tex]m_{AC} = \dfrac{3}{4}[/tex]

Since the diagonals of a square are perpendicular bisectors, the slope of diagonal BD is the negative reciprocal of the slope of diagonal AC:

[tex]m_{BD} = -\dfrac{1}{m_{AC}} = -\dfrac{1}{\frac{3}{4}} = -\dfrac{4}{3}[/tex]

Now, substitute the slope of BD and the coordinates of vertex B(4, -5) into the point-slope form of a linear equation to find the equation of diagonal BD:

[tex]\begin{aligned}y - y_B &= m_{BD}(x - x_B)\\\\y - (-5) &= -\dfrac{4}{3}(x - 4)\\\\y +5 &= -\dfrac{4}{3}(x - 4)\\\\3(y +5) &= -4(x - 4)\\\\3y+15&=-4x+16\\\\3y+15+4x-16&=-4x+16+4x-16\\\\4x+3y-1&=0\end{aligned}[/tex]

Therefore, the equation of diagonal BD is:

[tex]\Large\boxed{\boxed{4x+3y-1=0}}[/tex]

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