# Right triangle from axes

A line segment has its ends on the coordinate axes and forms with them a triangle of area equal to 36 square units. The segment passes through the point ( 5,2). What is the slope of the line segment?

Correct result:

k1 =  -0.015
k2 =  -10.705

#### Solution:

$S=36 \ \\ S=\dfrac{ ab }{ 2 } \ \\ ab=2 \cdot \ 36=72 \ \\ k=\dfrac{ 2-b }{ 5-a }=-\dfrac{ b }{ a } \ \\ \ \\ (2-b) a=-b \cdot \ (5-a) \ \\ (2-72/a) a=-72/a \cdot \ (5-a) \ \\ \ \\ (2a-72) a=-72 * (5-a) \ \\ \ \\ (2a-72) a=-72 \cdot \ (5-a) \ \\ 2a^2 -144a +360=0 \ \\ \ \\ p=2; q=-144; r=360 \ \\ D=q^2 - 4pr=144^2 - 4\cdot 2 \cdot 360=17856 \ \\ D>0 \ \\ \ \\ a_{1,2}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 144 \pm \sqrt{ 17856 } }{ 4 }=\dfrac{ 144 \pm 24 \sqrt{ 31 } }{ 4 } \ \\ a_{1,2}=36 \pm 33.40658617698 \ \\ a_{1}=69.40658617698 \ \\ a_{2}=2.5934138230199 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 2 (a -69.40658617698) (a -2.5934138230199)=0 \ \\ \ \\ \ \\ b_{1}=72/a_{1}=72/69.4066 \doteq 1.0374 \ \\ b_{2}=72/a_{2}=72/2.5934 \doteq 27.7626 \ \\ k_{1}=- \dfrac{ b_{1} }{ a_{1} }=- \dfrac{ 1.0374 }{ 69.4066 }=-0.015$

Checkout calculation with our calculator of quadratic equations.

$k_{2}=- \dfrac{ b_{2} }{ a_{2} }=- \dfrac{ 27.7626 }{ 2.5934 }=-10.705$

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Tips to related online calculators
For Basic calculations in analytic geometry is helpful line slope calculator. From coordinates of two points in the plane it calculate slope, normal and parametric line equation(s), slope, directional angle, direction vector, the length of segment, intersections the coordinate axes etc.
Looking for help with calculating roots of a quadratic equation?
Need help calculate sum, simplify or multiply fractions? Try our fraction calculator.
Pythagorean theorem is the base for the right triangle calculator.
See also our trigonometric triangle calculator.

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