Vertex A of square ABCD is at (22, 10). BD lies on 4y - 3x = 24. B closer than D to origin. Find....

[Ayries]

7. The vertex A of a square ABCD is at the point (22, 10). The diagonal BD has equation 4y - 3x = 24, and the vertex B is nearer to the origin than D.

(i) Calculate the coordinates of the centre of the square.

(ii) Calculate the coordinates of B and C.

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I'm having trouble with part (ii), finding the coordinates of vertex B. Please help. Thank you! ?

 

[Ayries]

Pls help me with question 7 (ii) , thanks !!

 

[Deleted member 4993]

7. The vertex A of a square ABCD is at the point (22, 10). The diagonal BD has equation 4y - 3x = 24, and the vertex B is nearer to the origin than D.

(i) Calculate the coordinates of the centre of the square.

(ii) Calculate the coordinates of B and C.

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I'm having trouble with part (ii), finding the coordinates of vertex B.

Did you solve part I? If you did please share your work - we will start from there.

If you did not, start with trying to sketch the square (including the x-y axis) according to the description given.

 

[Ayries]

Yes , here's part 1

 

[pka]

7. The vertex A of a square ABCD is at the point (22, 10). The diagonal BD has equation 4y - 3x = 24, and the vertex B is nearer to the origin than D.

(i) Calculate the coordinates of the centre of the square.

(ii) Calculate the coordinates of B and C.

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I'm having trouble with part (ii), finding the coordinates of vertex B.

Here are useful facts. In a square the diagonals are perpendicular and bisect each other. therefore you know the slope of \(\displaystyle \overleftrightarrow {AB}\)

Also, if \(\displaystyle P(x_p,y_p)\) is a point and \(\displaystyle \ell:~ax+by+c=0\) is a line then the distance from\(\displaystyle P\text{ to }\ell\text{ is }\dfrac{|a\cdot x_p+b\cdot y_p+c|}{\sqrt{a^2+b^2}} \)
Find distance \(\displaystyle A\) is from the given diagonal: \(\displaystyle 3x-y+24=0\).