ULYSESAUTORidn
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find the distance from the point (6,-5) to the line 2x -3y-2=0​

Sagot :

AGboiiidn

Answer:

[tex]\frac{25\sqrt{13}}{13}[/tex]

Step-by-step explanation:

Given:

  • 2x-3y-2 = 0
  • (6,-5)

So ,

  • a = 2 , b = -3 , c = -2
  • x1 = 6 , y1 = -5

Solution:

  • [tex]d = \frac{ax_1+by_1+C}{\sqrt{a²+b²}}[/tex]

  • [tex]d = \frac{(2)(6)+(-3)(-5)+(-2)}{\sqrt{2²+(-3)²}}[/tex]

  • [tex]d = \frac{12+15-2}{\sqrt{4+9}}[/tex]

  • [tex]d = \frac{25}{\sqrt{13}}[/tex]

  • [tex]d = \frac{25}{\sqrt{13}}×\frac{\sqrt{13}}{\sqrt{13}}[/tex]

  • [tex]d = \frac{25\sqrt{13}}{13}[/tex]

[tex] \\ [/tex]

Therefore , the distance is [tex] \frac{25\sqrt{13}}{13}[/tex]

Subcribe198idn

Answer:

✏PROBLEM:

Find the distance from the point (6,-5) to the line 2x -3y-2=0?

✏ANSWER:

[tex] \frac{25}{ \sqrt{13} } \: units [/tex]

✏SOLUTION:

IN THE PICTURE PLEASE

View image Subcribe198