Math Snap

STEP 1

Assumptions1. Triangle $CAB$ is a right-angled triangle.
. The coordinates of point $A$ are $(-12,7)$.
3. The coordinates of point $B$ are $(-9,b)$.
4. The coordinates of point $C$ are $(c,11)$.
5. The coordinates of point $$ are $(p,1)$.
6. The line $AC$ intersects the $y$-axis at $Q$ such that $AQ=QC$.
7. Point $$ lies on line $AB$ and $AB$ is perpendicular to $AC$.

STEP 2

Since $AB$ is perpendicular to $AC$, the slopes of $AB$ and $AC$ are negative reciprocals of each other. Let's find the slope of $AC$.
$$lope_{AC} = \frac{y2 - y1}{x2 - x1} = \frac{11 -7}{c - (-12)} = \frac{4}{c +12}$$

STEP 3

The slope of $AB$ is the negative reciprocal of the slope of $AC$.
$$lope_{AB} = -\frac{1}{lope_{AC}} = -\frac{c +12}{}$$

STEP 4

We can also find the slope of $AB$ using the coordinates of $A$ and $B$.
$$lope_{AB} = \frac{b -7}{-9 - (-12)} = \frac{b -7}{3}$$

STEP 5

Setting the two expressions for $lope_{AB}$ equal to each other, we can solve for $c$.
$$-\frac{c +12}{4} = \frac{b -7}{3}$$

STEP 6

olving for $c$ gives us$$c = -4 \times \frac{b -}{3} -12$$

STEP 7

Since $AQ = QC$, the $x$-coordinate of $Q$ is $0$ (as it lies on the $y$-axis) and the $y$-coordinate of $Q$ is the average of the $y$-coordinates of $A$ and $C$.
$$y_Q = \frac{7 +11}{2} =9$$So, $Q = (0,9)$.

STEP 8

Substituting $b =$ into the equation from6, we can solve for $c$.
$$c = -4 \times \frac{ -7}{3} -12 = -4$$

STEP 9

We can find the equation of line $AB$ using the point-slope form of the equation of a line, $y - y = m(x - x)$, where $m$ is the slope and $(x, y)$ is a point on the line.
$$y -7 = \frac{2}{3}(x +12)$$

STEP 10

Substituting $y =$ and solving for $x$ gives us $p$.
$$-7 = \frac{2}{3}(p +12)$$$$p = \frac{3( -7)}{2} -12 = -15$$

The ratio $APPB$ is equal to the ratio of the distances from $A$ to $$and from$$ to $B$.
APPB = \frac{\sqrt{(- - (-15))^ + (7 -)^}}{\sqrt{(-15 - (-9))^ + ( -9)^}} = \frac{\sqrt{9 +36}}{\sqrt{36 +64}} = \frac{\sqrt{45}}{\sqrt{100}} = \frac{3}{}(a) The value of $c$ is $-4$.
(b) The coordinates of $Q$ are $(0,9)$.
(c) The values of $b$ and $p$ are $9$ and $-15$, respectively.
(d) The ratio $APPB$ is $3$.