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3 years ago

Does the point (-2, -5) satisfy the equation y = x − -3?

Mathematics

2 answers:

Fed [463]3 years ago

6 0

Answer:

Yes

Step-by-step explanation:

To determine if (- 2, - 5) is a solution to the equation.

Substitute x = - 2 into the right side of the equation and if the value obtained is equal to the y- coordinate of the point then it is a solution.

y = - 2 - 3 = - 5 ← hence (- 2, - 5) is a solution to the equation

Verizon [17]3 years ago

4 0

No because (-2, -5) is negative and the y coordinate needs to be positive.

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3 years ago

P (2, 3) and Q (6, 10) are two points on the Cartesian plane.

lara31 [8.8K]

Step-by-step explanation:

Given:

$(x_1, \: y_1 ) =(2, \:3)\:\:\&\:\: (x_2, \: y_2) =(6, \:10)$

$l(PQ) = \sqrt{ {(x_1 - x_2)}^{2} + {(y_1 - y_2)}^{2}} \\ \\ = \sqrt{ {(2 - 6)}^{2} + {(3 - 10)}^{2}} \\ \\ = \sqrt{ {( - 4)}^{2} + {( - 7)}^{2}} \\ \\ = \sqrt{16 + 49} \\ \\ = \sqrt{65} \\ \\ \therefore \purple{ \boxed{ l(PQ) = 8.06 \: units}}\\\\$

Let S be the mid-point of PQ.

$\therefore S = \{\frac{x_1+x_2}{2}, \:\:\frac{y_1+y_2}{2}\} \\\\= \{\frac{2 +6}{2}, \:\:\frac{3+10}{2}\} \\\\= \{\frac{8}{2}, \:\:\frac{13}{2}\} \\\\\therefore S= \{4, \:\:6.5\} \\\\$

Equation of line PQ is given as:

$\frac{y-y_1}{y_1 - y_2}=\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-3}{3 - 10}=\frac{x-2}{2 - 6} \\\\\therefore \frac{y-3}{-7}=\frac{x-2}{-4} \\\\\therefore \frac{y-3}{7}=\frac{x-2}{4} \\\\\therefore 4(y-3)=7(x-2)\\\\\therefore 4y-12=7x-14\\\\\therefore 7x-14 - 4y + 12=0\\\\\red{\boxed {\therefore 7x-4y - 2 =0}}$

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3 years ago

Read 2 more answers