# Can anyone help with this geometry question (finding the point with provides for a ratio for the areas of two triangles)? **Note - I'm struggling with the last part of this problem (with the ratios of the areas of the triangles)

Figure 1 on the right shows a graph where the curve ℓ represents the function y ＝ x 2 and the curve m represents the function y＝ax 2 （a＜0）. The point O represents the origin of the graph. Point A lies on curve m. The x-coordinate of point A is negative, and its y-coordinate is －1. Point B lies on curve ℓ, and has the same x-coordinate as point A. Let point P be a point on curve ℓ with a positive x-coordinate. Answer the following questions.

〔Question 1 〕 Consider the case where the values of the y-coordinates of points P and B are equal. Answer（１）and（２）.

（１） Find the value of a when the x-coordinate of point P is 4.

（２） Find the equation of the line which passes through points A and P when the value of a is -1/9. 〔Question 2 〕 Figure 2 on the right shows the case in Figure 1 where the value of a is -1/4. Let point C be a point on curve m with its x-coordinate 4. Let point D be a point on the y-axis with the same y-coordinate as point C. Connect points A and B, points A and P, points B and P, points C and P, points C and D, and points D and P. Find the coordinates of point P when the area of triangle DCP is four times larger than the area of triangle BAP.

Thanks!

• You stated that you're struggling with the last part, i.e., the ratios of the areas of the triangles. Please tell us, by updating the question text, what you've tried so far and, in particular, what you're having trouble with. Thanks. – John Omielan Aug 8 at 0:15
• Can you find length of $AB$ and $DC$? – Vasya Aug 8 at 0:55
• I'm not sure how to get started - I've found the lengths of $AB$ (length: 5) and $DC$ (length: 4) and I've tried trial and error for point B, but I can't seem to find a solution. – mathinjpm Aug 8 at 1:03
• Let $A_x = - k$, a known quantity. Then the altitude of ABP using AB as base is k + 4. – Mick Aug 8 at 4:46

So you found that $$AB=5$$ and $$DC=4$$. Let $$P$$ have coordinates $$(x, x^2)$$. Then the area of BAP $$A_{\triangle BAP}= 0.5 \cdot 5 \cdot (x+2)$$. Similarly, $$A_{\triangle DCP}= 0.5 \cdot 4 \cdot (x^2+4)$$. Therefore we have this equation to find $$x$$: $$10 \cdot (x+2)=2 \cdot (x^2+4)$$ Can you finish?
• Thank you - the answer is correct! Just one question, so I completely understand. What does the $x+2$ (or $x^2+4$) term represent? i.e. Why is there a $+2$ and a $+4$? – mathinjpm Aug 8 at 20:33
• @mathinjpm: $x+2$ and $x^2+4$ are heights of those triangles (because one side of each triangle is parallel either axis $x$ or $y$, the height will be the difference in either $x$ or $y$ coordinates). let me know if you need more explanation. – Vasya Aug 8 at 23:26