The diagram shows a square ABCD with a point inside it which is 1 unit from D, 2 units from A and 3 units from B. Let angle APD= theta . (a) Show that theta = (3pi)/4 (b Show that if P is outside the square , then theta = pi/4 () Is the situation possible if P is 3 units from C instead of from B?

The diagram shows a square ABCD with a point inside it which is 1 unit from D, 2 units from A and 3 units from B. Let angle APD= theta . (a) Show that theta = (3pi)/4 (b Show that if P is outside the square , then theta = pi/4 () Is the situation possible if P is 3 units from C instead of from B?

Step-by-step explanation:

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step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?