Qn06: (a). Plot the graph; y = x2 + x - 2 (b). Verify that the perpendicular bisector of ABC are concurrent. mAB = 5cm ; mBC = 6cm ; mAC = 7cm

Mathematics
Qn06: (a). Plot the graph; y = x2 + x - 2 (b). Verify that the perpendicular bisector of ABC are concurrent. mAB = 5cm ; mBC = 6cm ; mAC = 7cm

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Q2(a)

Step 1: Rewrite the equation piecewise.

y=∣x∣−3={x−3x≥0−x−3x<0y = |x| - 3 = \begin{cases} x - 3 & x \geq 0 \\ -x - 3 & x < 0 \end{cases}

Step 2: Calculate key points (in cm units if scaled, but graph is unitless).

&x = -4,\ y = |-4| - 3 = 1 \\ &x = -3,\ y = 0 \\ &x = -2,\ y = -1 \\ &x = -1,\ y = -2 \\ &x = 0,\ y = -3 \\ &x = 1,\ y = -2 \\ &x = 2,\ y = -1 \\ &x = 3,\ y = 0 \\ &x = 4,\ y = 1 \end{align*} $$ Step 3: Plot points and connect: V-shape symmetric about y-axis, vertex at $(0, -3)$, slopes $\pm 1$, x-intercepts at $x = \pm 3$. **Q2(b)** Step 1: Assign coordinates (units in cm). Place B at $(0,0)$, C at $(6,0)$ since BC = 6 cm. Let A at $(p, q)$. Distance AB = 5 cm: $$ p^{2} + q^{2} = 25 $$ AC = 7 cm: $$ (p - 6)^{2} + q^{2} = 49 $$ Subtract equations: $$ (p - 6)^{2} + q^{2} - p^{2} - q^{2} = 24 $$ $$ p^{2} - 12p + 36 - p^{2} = 24 $$ $$ -12p + 36 = 24 $$ $$ -12p = -12 $$ $$ p = 1 $$ Substitute into first: $$ 1 + q^{2} = 25 $$ $$ q^{2} = 24 $$ $$ q = 2\sqrt{6} $$ (positive for triangle above x-axis). A$(1, 2\sqrt{6})$, B$(0,0)$, C$(6,0)$ cm. Step 2: Perp bisector of BC. Midpoint: $(3, 0)$ cm. Perp to x-axis: vertical line $$ x = 3 $$ Step 3: Perp bisector of AB. Midpoint: $\left( \frac{1+0}{2}, \frac{2\sqrt{6}+0}{2} \right) = (0.5, \sqrt{6})$ cm. Slope AB: $\frac{2\sqrt{6}-0}{1-0} = 2\sqrt{6}$. Perp slope: $-\frac{1}{2\sqrt{6}} = -\frac{\sqrt{6}}{12}$. Equation: $$ y - \sqrt{6} = -\frac{\sqrt{6}}{12} (x - 0.5) $$ Step 4: Intersection with $x=3$ cm. $$ y - \sqrt{6} = -\frac{\sqrt{6}}{12} (3 - 0.5) = -\frac{\sqrt{6}}{12} \cdot 2.5 = -\frac{\sqrt{6}}{12} \cdot \frac{5}{2} = -\frac{5\sqrt{6}}{24} $$ $$ y = \sqrt{6} - \frac{5\sqrt{6}}{24} = \frac{24\sqrt{6} - 5\sqrt{6}}{24} = \frac{19\sqrt{6}}{24} $$ Circumcenter O$\left(3, \frac{19\sqrt{6}}{24}\right)$ cm. Step 5: Perp bisector of AC. Midpoint: $\left( \frac{1+6}{2}, \frac{2\sqrt{6}+0}{2} \right) = \left( \frac{7}{2}, \sqrt{6} \right)$ cm. Slope AC: $\frac{0 - 2\sqrt{6}}{6-1} = -\frac{2\sqrt{6}}{5}$. Perp slope: $\frac{5}{2\sqrt{6}} = \frac{5\sqrt{6}}{12}$. Equation: $$ y - \sqrt{6} = \frac{5\sqrt{6}}{12} \left( x - \frac{7}{2} \right) $$ Step 6: Substitute $x=3$ cm. $$ y - \sqrt{6} = \frac{5\sqrt{6}}{12} \left(3 - \frac{7}{2}\right) = \frac{5\sqrt{6}}{12} \left(-\frac{1}{2}\right) = -\frac{5\sqrt{6}}{24} $$ $$ y = \sqrt{6} - \frac{5\sqrt{6}}{24} = \frac{19\sqrt{6}}{24} $$ cm. Passes through O. **All perpendicular bisectors are concurrent at O. Verified.**
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