Solution of Area on the Coordinate Plane (Questions published on Dec 8, 2025)
Area on the Coordinate Plane
Full Step-by-Step Solutions (Problems 1–15)
Working conventions / reminders
- For an equation of the form \(|x-h|+|y-k|=R\) the graph is a diamond (rotated square) with area \(2R^2\).
- For \(a|x-h| + b|y-k| = c\): horizontal half-width \(=c/a\), vertical half-height \(=c/b\). Diagonals: \(d_x=2c/a,\ d_y=2c/b\). Area \(=\dfrac{d_x d_y}{2}.\)
- Area of a circle: \(A=\pi r^2\). Distance from \((x_0,y_0)\) to line \(Ax+By+C=0\) is \(\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}\).
Solutions
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Problem 1.
\(y = -|x+4| + 7\) and \(y = |x+4| - 5\). Find the enclosed area.1. Set the two expressions equal to find intersection points: \[ -|x+4|+7 = |x+4|-5 \Longrightarrow 12 = 2|x+4| \Longrightarrow |x+4|=6. \] So \(x+4 = \pm 6 \Rightarrow x = -10,\; x = 2.\)2. At those \(x\)-values, compute \(y\): \(y = |x+4|-5 = 6-5 = 1\). So intersections are \((-10,1)\) and \((2,1)\).3. The top vertex of the top curve \(y=-|x+4|+7\) is at \(x=-4\), \(y=7\). The bottom vertex of the lower curve \(y=|x+4|-5\) is at \((-4,-5)\).4. Horizontal diagonal length \(=2-(-10)=12\). Vertical diagonal length \(=7-(-5)=12\).Area \(=\dfrac{12\cdot12}{2}=72.\) \(\boxed{72}\) -
Problem 2.
\(y = 12 - |x-6|\). Find the area of the region enclosed by the V.1. The vertex is at \((6,12)\). Find x-intercepts where \(y=0\): \[ 12 - |x-6| = 0 \Rightarrow |x-6| = 12 \Rightarrow x = -6,\; x=18. \]2. This forms an isosceles triangle with base length \(18-(-6)=24\) and height \(12\).Area \(=\tfrac12\cdot 24\cdot 12 = 144.\) \(\boxed{144}\) -
Problem 3.
\(|x-8| + |y-1| = 10.\)This is a diamond centered at \((8,1)\) with parameter \(R=10\). Formula: area \(=2R^2\).Area \(=2\cdot 10^2 = 200.\) \(\boxed{200}\) -
Problem 4.
\(3|x| + 2|y| = 18.\)1. When \(y=0\): \(3|x|=18\Rightarrow|x|=6\) so x-intercepts \((\pm6,0)\) ⇒ horizontal diagonal \(=12\).2. When \(x=0\): \(2|y|=18\Rightarrow|y|=9\) so y-intercepts \((0,\pm9)\) ⇒ vertical diagonal \(=18\).Area \(=\dfrac{12\cdot18}{2}=108.\) \(\boxed{108}\) -
Problem 5.
\(4|x-2| + |y+3| = 20.\) Find the absolute difference between the diagonals.1. Horizontal diagonal: set \(y=-3\) so \(|y+3|=0\Rightarrow 4|x-2|=20\Rightarrow |x-2|=5\). Solutions \(x=-3,7\). Horizontal diagonal \(=7-(-3)=10\).2. Vertical diagonal: set \(x=2\) so \(|x-2|=0\Rightarrow |y+3|=20\Rightarrow y+3=\pm20\Rightarrow y=17\) or \(-23\). Vertical diagonal \(=17-(-23)=40\).Absolute difference \(=|40-10|=30.\) \(\boxed{30}\) -
Problem 6.
\((x-4)^2 + (y-9)^2 = 25.\) Area?Radius \(r=\sqrt{25}=5\). Area of circle \(A=\pi r^2=\pi\cdot5^2=25\pi\).\(\boxed{25\pi}\) -
Problem 7.
\(|x-a| + |y-2| = b\) has area \(72\). Find \(b\).Area of such a diamond \(= 2b^2\). So \(2b^2 = 72 \Rightarrow b^2 = 36 \Rightarrow b = 6\).\(\boxed{6}\) -
Problem 8.
\(6|x+1| + 3|y-5| = k\) and area \(=54\). Find \(k\).Use formula: for \(a|x-h|+b|y-k|=c\), area \(= \dfrac{2c^2}{ab}\).Here \(a=6,\;b=3,\;c=k\). So area \(=\dfrac{2k^2}{6\cdot3}=\dfrac{2k^2}{18}=\dfrac{k^2}{9}.\)Set \(\dfrac{k^2}{9}=54 \Rightarrow k^2 = 486 = 81\cdot6 \Rightarrow k = \sqrt{486} = 9\sqrt{6}\).\(\boxed{9\sqrt{6}}\) -
Problem 9.
\(7|x| + 24|y| = 84.\) A circle is inscribed — find its area.Consider the first-quadrant boundary line: \(7x + 24y = 84\). The inscribed circle will be tangent to that line and centered at the origin \((0,0)\).Distance from origin to line \(7x+24y-84=0\) is \[ r = \frac{| -84 |}{\sqrt{7^2 + 24^2}} = \frac{84}{25}. \]The distance from the center to each side is the radius of the inscribed circle. For the line \(7|x| + 24|y| = 84\), one side in the first quadrant is \(\displaystyle 7x + 24y = 84\). The distance from the origin to this line is: \[ r = \frac{84}{\sqrt{7^{2} + 24^{2}}} = \frac{84}{25} = \frac{84}{25}. \] Thus the area of the inscribed circle is: \[ \text{Area} = \pi r^{2} = \pi \left(\frac{84}{25}\right)^{2} = \frac{7056}{625}\pi. \]
.\(\boxed{\dfrac{7056\pi}{625}}\) -
Problem 10.
\((x-3)^2 + (y+4)^2 = 32.\) A square is inscribed in the circle — find the area of the square.Radius \(r=\sqrt{32}=4\sqrt{2}\). Diameter \(=2r=8\sqrt{2}\) equals the square's diagonal.Area of square \(= \dfrac{\text{(diagonal)}^2}{2} = \dfrac{(8\sqrt{2})^2}{2} = 64.\)\(\boxed{64}\) -
Problem 11.
\(y = |x+10| - 3\) and \(y = -|x+10| + 9.\) Find enclosed area.Set equal: \(|x+10|-3 = -|x+10| + 9 \Rightarrow 2|x+10| = 12 \Rightarrow |x+10| = 6.\)So \(x+10 = \pm 6 \Rightarrow x = -16,\; -4\). At these points \(y = 6 - 3 = 3\). Intersection points \((-16,3)\) and \((-4,3)\).Vertex of top curve is \((-10,9)\). Vertex of bottom curve is \((-10,-3)\). Vertical diagonal \(=12\). Horizontal diagonal \(=12\).Area \(=\dfrac{12\cdot12}{2}=72.\) \(\boxed{72}\) -
Problem 12.
\(|x-2| + 3|y| = 15.\) Find the area.Horizontal half-width \(=15\) ⇒ diagonal \(=30\). Vertical half-height \(=5\) ⇒ diagonal \(=10\).Area \(=\dfrac{30\cdot10}{2} = 150.\) \(\boxed{150}\) -
Problem 13.
\(2|x-1| + 5|y+6| = p\). Diagonals differ by 6 — find \(p\).Horizontal diagonal \(d_x = p\).Vertical diagonal \(d_y = \frac{2p}{5}\).\(|d_x - d_y| = 6 \Rightarrow \left| p - \frac{2p}{5} \right| = 6 \Rightarrow \frac{3p}{5} = 6 \Rightarrow p = 10.\)\(\boxed{10}\) -
Problem 14.
\((x+12)^2 + (y-1)^2 = 18.\) Area of a regular hexagon inscribed in this circle?Radius \(r=\sqrt{18}=3\sqrt{2}\). For a regular hexagon, side length \(s=r\).Area \(= \dfrac{3\sqrt{3}}{2} r^2.\)\[ A = \dfrac{3\sqrt{3}}{2}\cdot 18 = 27\sqrt{3}. \]\(\boxed{27\sqrt{3}}\) -
Problem 15.
\(|x| + |y| = 14\). A rectangle with sides parallel to axes is inscribed. Maximize its area.Use first-quadrant point with \(x+y=14\).Area \(=4xy\).Max when \(x=y=7\).Maximum area \(=196.\) \(\boxed{196}\)
Answer summary (quick reference)
- 72
- 144
- 200
- 108
- 30
- \(25\pi\)
- 6
- \(9\sqrt{6}\)
- \(\dfrac{7056\pi}{625}\)
- 64
- 72
- 150
- 10
- 27\(\sqrt{3}\)
- 196
Hope it helps.