QuestionIn triangle $CAB$ , find $c$ , coordinates of $Q$ , values of $b$ and $p$ , and ratio $AP:PB$ . Points: $A(-12,7)$ , $B(-9,b)$ , $C(c,11)$ , $P(p,1)$ .

Studdy Solution

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 $p$ .
$-7 = \frac{2}{3}(p +12)$ $p = \frac{3( -7)}{2} -12 = -15$

STEP 11

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 $3$ .

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