Question

Distance of point (-3, 4) from the origin is ..​

Answer 1

Answer:

1. let the point be a₍-3,4₎ and O₍0,0₎

∴ by distance formula√ $(x2-x1)2+(y2-y1)2$

√(3-0)*2+(4-0)*2

√9+16

√25

∴5 units

Explanation:

Answer 2

$\underline{\underline{\textsf{\maltese\:\: {\red{Given :}}}}}$

• Points of origin = (x₁ , y₁) = (0,0)

• Another Point = (x₂ , y₂) = (-3,4)

$\underline{\underline{\textsf{\maltese\:\: {\red{Diagram :}}}}}$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\vector(0,1){6}} \put(0,0){\vector(0,-1){6}} \put(0,0){\vector(1,0){6}} \put(0,0){\vector(-1,0){6}} \put(3.9,-.3){\bf{+ X - Axis}} \put(-5.9,-.3){\bf{- X - Axis}} \put(-1,-6.5){\bf{- Y - Axis}} \put(-1,6.5){\bf{+ Y - Axis}}\put(3,3){\bf + , +} \put(-3,3){\bf{ - , +}}\put(3,-3){\bf + , -}\put(-3,-3){\bf - , -} \put(.1,-.3){\bf{(0,0) Origin}}\put(-1,8){\bf Cartesian Plane}\put(5.5,-7.5){\framebox(2.7,.7)} \put(5.5,-7.5){\bf@ BeBrainliest}\qbezier(0,0)(0,0)(-3,4) \put(-3,4.2){\bf(-3,4)}\put(-3,4){\circle*{.1}} \put(0,0){\circle*{.1}}\qbezier(-1,7.7)(-1,7.7)(1.9,7.7)\end{picture}$

$\underline{\underline{\textsf{\maltese\:\: {\red{Solution :}}}}}$

In order to find the distance between two points we need to use the Distance Formula.

D = √ (x₂ - x₁)² + (y₂ - y₁)²

D = √ (-3 - 0)² + (4 - 0)²

D = √(-3)² + 4²

D = √8 + 16

D = √25

D = 5 units

∴ Distance of point (-3,4) from the origin is ${\underline{\textbf{5 units}}}$.