# The vertices of trianglePQR are P(1,3), Q(5,4) & R(5,15). Find the length of PR, area of TrianglePQR & length of per. line drown fm Q to PR.

crmhaske | College Teacher | (Level 3) Associate Educator

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This might look confusing because it isn't possible to show you any diagrams.

To solve the length of PR first find how far over (x), and up (y) R is from P.

x = xR-xP = 5 - 1 = 4
y = yR=yP = 15-3 = 12

Use pythagorean theorm to solve for PR (the hypotenuse):

PR = sqrt(x^2 + y^2) = sqrt(4^2 + 12^2) = 12.6

To solve the area we first need the height, which is the perpendicular line from Q to PR.  First we need the lengths of the other two sides PQ, and RQ.  We solve the same way as we did to find PR:

xPQ = xQ-xP = 4
yPQ = yQ-yP = 1

PQ = sqrt(4^2 + 1^2) = 4.1

xRQ = 0
yRQ = yR = yQ = 15-4 = 11

Using the cosine law of a non-right triangle (c^2 = a^2 + b^2 -2bc(cos[theta])) we can solve for angle PRQ:

4.1^2 = 12.6^2 + 11^2 - 2(12.6)(11)cos[theta]
theta = 18.45 deg

Using sine law in a right triangle we can find the height (h):

sin(18.45) = h/11 --> h = 11*sin(18.45) = 3.5

Finally, to find the area of the triangle:

A = 0.5*b*h = 0.5*PR*h = 0.5*12.6*3.5 = 22

neela | High School Teacher | (Level 3) Valedictorian

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To find the area of the triangle ABC whose vertics are P(1,3),Q(5,4) and R(5,15). By distance formula ,

PQ^2 = r^2 = (5-1)^2+(4-3)^2 = 16+1 = 17

PR^2 = q^2 = (5-1)^2+(15-3)^2 = 16+144 = 160

QR^2 =p ^2 = (5--5)^2+(15-4)^2 = 0+121 =121

Therefore  PR =  q = sqrt160 = 4sqrt10.

The equation of PR:   y - Px = (Ry-Py)/(Rx-Px) { x- Px}

y-3 = (15-3)/(5-1) {x-1}

y-3 = 3(x-1)

3x-y = 0

The length of perpendicular from (x1,y1) ax+bx+c = 0 is  (ax1+by1+c)/sqrt(a^2+b^2)

Therefore Length of perpendicular from Q(5,4) to 3x-y+0 = 0 is

= | 3(5)-1(4))|/sqrt((3^2+(-1)^2) = 11/sqrt10.

Therefore area of the triangle = (1/2) (PR) *(11/sqrt10) =

(1/2)(4sqrt10)(11/sqrt10) = 22 sq units

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To find the length of PR, we'll use the distance formula:

PR= sqrt[(xR-xP)^2 + (yR-yP)^2]

PR = sqrt[(5-1)^2+(15-3)^2]

PR = sqrt(16+144)

PR = sqrt 160

PR = sqrt 16*10

PR = 4*sqrt 10

Let's calculate the length of the segment QP

QP = SQRT[(xP-xQ)^2+(yP-yQ)^2]

QP = sqrt[(1-5)^2+(3-4)^2]

QP = sqrt(16+1)

QP = sqrt 17

Now, let's calculate QR

QR = sqrt [(5-5)^2+(15-4)^2]

QR = sqrt 11^2

QR = 11

Area of the triangle will be calculated with Heron formula:

A = sqrt[p(p-QP)(p-QR)(p-PR)]

where p = (QR+QP+PR)/2

To calculate the height QM, we'll have to find out the equation of PR , using the standard form y = mx+n, where m is the slope of PR. After that, we'll consider the constraint the 2 lines are perpendicular if and only if the product of  their slopes is -1.

To find the equation of PR, we'll consider the formula:

(xR-xP)/(x-xP) = (yR-yP)/(y-yP)

(5-1)/(x-1) = (15-3)/(y-3)

4/(x-1) = 12/(y-3)

We'll divide by 4 both sides:

1/(x-1) = 3/(y-3)

We'll cross multiply:

3x-3 = y-3

We'll add 3 both sides:

3x = y

So the slope of PR is m1 = 3

The slope of QM is m2 = -1/3.

The equation of QM is:

y - yQ = (-1/3)(x-xQ)

y-4 = (-1/3)(x-5)

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