Graphing linear equations

randRangeNonZero( -10, 10 ) randRangeNonZero( -5, 5 ) B * YINT 0 randRange( 2, 6 )

Graph the following equation:

expr([ "*", B, "y" ]) = C

Solve for y to convert expr([ "*", B, "y" ]) = C to slope-intercept form.

Divide both sides by B:

\qquad y = \color{GREEN}{YINT}

The y-intercept is YINT, and this equation represents a line where all of the points have y = YINT.

Let's pick some points where y = YINT. (-X_EXAMPLE, YINT) and (X_EXAMPLE, YINT) are both good choices.

style({ stroke: GREEN, fill: GREEN }, function() { graph.exampleA = circle( [ -1 * X_EXAMPLE, YINT ], 0.2 ).toBack(); graph.exampleB = circle( [ X_EXAMPLE, YINT ], 0.2 ).toBack(); });

We can see that these points line up to form a horizontal line that crosses the y-axis at (0, YINT).

style({ stroke: BLUE, strokeWidth: 2, strokeDasharray: "." }, function() { plot( function( x ) { return YINT; }, [ -11, 11 ] ).toBack(); });

randRangeNonZero( -10, 10 ) randRangeNonZero( -5, 5 ) B * XINT randRange( 2, 6 ) randRange( 2, 6 )

Graph the following equation:

expr([ "*", B, "x" ]) = C

Notice that there is no y in this equation. Try solving for x and see what happens.

Divide both sides by B:

\qquad x = \color{GREEN}{XINT}

This equation represents a line where all of the points have x = XINT.

Let's pick some points where x = XINT. (XINT, -Y_EXAMPLE), (XINT, Y_EXAMPLE_2) and (XINT, Y_EXAMPLE) are all good choices.

style({ stroke: GREEN, fill: GREEN }, function() { graph.exampleA = circle( [ XINT, -1 * Y_EXAMPLE ], 0.2 ).toBack(); graph.exampleB = circle( [ XINT, Y_EXAMPLE ], 0.2 ).toBack(); graph.exampleA = circle( [ XINT, Y_EXAMPLE_2 ], 0.2 ).toBack(); });

We can see that these points line up to form a vertical line that crosses the x-axis at (XINT, 0).

style({ stroke: BLUE, strokeWidth: 2, strokeDasharray: "." }, function() { line([XINT, -11], [XINT, 11]).toBack(); });

[ graph.pointA.coord, graph.pointB.coord ]

return abs(guess[0][0] - XINT) < 0.001 && abs(guess[1][0] - XINT) < 0.001;

graph.pointA.setCoord( guess[0] ); graph.pointB.setCoord( guess[1] ); graph.line1.transform( true );

reduce( randRangeNonZero( -5, 5 ), randRange( 1, 5 ) ) randRangeNonZero( max( -10, -10 - SLOPE_FRAC[0] ), min( 10, 10 - SLOPE_FRAC[0] ) ) SLOPE_FRAC[0] / SLOPE_FRAC[1] SLOPE === 1 ? "" : ( SLOPE === -1 ? "-" : fraction( SLOPE_FRAC[0], SLOPE_FRAC[1], true, true ) ) randRangeNonZero( -3, 3 ) SLOPE_FRAC[0] * -MULT SLOPE_FRAC[1] * MULT SLOPE_FRAC[1] * YINT * MULT random() < 0.3

Graph the following equation:

expr([ "+", [ "*", A, "x" ], [ "*", B, "y" ] ]) = C

y = PRETTY_SLOPE x + YINT

Convert expr([ "+", [ "*", A, "x" ], [ "*", B, "y" ] ]) = C to slope-intercept form by solving for y.

Add abs( A )x to both sides:

Subtract abs( A )x from both sides:

\qquad expr( [ "*", B, "y" ] ) = expr([ "+", [ "*", -A, "x" ], C ])

Divide both sides by B:

\qquad y = expr([ "+", "\\dfrac{" + expr([ "*", -A, "x" ]) + "}{" + B + "}", "\\dfrac{" + C + "}{" + B + "}" ])

\qquad y = \color{purple}{PRETTY_SLOPE} x \color{GREEN}{+ YINT}

The y-intercept is YINT and the slope is decimalFraction( SLOPE, true, true ).

The y-intercept is YINT, so the line must pass through the point (0, YINT).

style({ stroke: GREEN, fill: GREEN }, function() { graph.yint = circle( [ 0, YINT ], 0.2 ).toBack(); });

The slope is decimalFraction( SLOPE, true, true ). Remember that the slope tells you rise over run. For every step we take we must move abs( SLOPE_FRAC[0] ) positions down (because it's negative)up. For every step we take we must move abs( SLOPE_FRAC[0] ) position down (because it's negative)up. For every step we take we must also move SLOPE_FRAC[1] positions to the right. For every step we take we must also move SLOPE_FRAC[1] position to the right.

style({ stroke: "purple", strokeWidth: 2, arrows: "->" }, function() { path([ [ 0, YINT ], [ 0, YINT + SLOPE_FRAC[0] ] ]).toBack(); path([ [ 0, YINT + SLOPE_FRAC[0] ], [ SLOPE_FRAC[1], YINT + SLOPE_FRAC[0] ] ]).toBack(); }); label( [ 0, YINT + SLOPE_FRAC[0] / 2 ], abs( SLOPE_FRAC[0] ) + " \\text{ " + ( SLOPE_FRAC[0] < 0 ? i18n._("down") : i18n._("up") ) + "} \\quad", "left", { color: "purple" } ); label( [ 0, YINT + SLOPE_FRAC[0] ], SLOPE_FRAC[1] + " \\text{ " + i18n._("right") + "}", ( SLOPE_FRAC[0] < 0 ? "below right" : "above right" ), { color: "purple" } ); graph.yint.toBack();

So the line must also pass through (SLOPE_FRAC[1], YINT + SLOPE_FRAC[0])

style({ stroke: BLUE, strokeWidth: 2, strokeDasharray: "." }, function() { plot( function( x ) { return x * SLOPE + YINT; }, [ -11, 11 ] ).toBack(); });

graphInit({ range: 11, scale: 20, axisArrows: "<->", tickStep: 1, labelStep: 1, gridOpacity: 0.05, axisOpacity: 0.2, tickOpacity: 0.4, labelOpacity: 0.5 }); label( [ 0, 11 ], "y", "above" ); label( [ 11, 0 ], "x", "right" ); addMouseLayer(); graph.pointA = addMovablePoint({ coord: [ -5, 5 ], snapX: 0.5, snapY: 0.5, normalStyle: { stroke: KhanUtil.BLUE, fill: KhanUtil.BLUE } }); graph.pointB = addMovablePoint({ coord: [ 5, 5 ], snapX: 0.5, snapY: 0.5, normalStyle: { stroke: KhanUtil.BLUE, fill: KhanUtil.BLUE } }); graph.line1 = addMovableLineSegment({ pointA: graph.pointA, pointZ: graph.pointB, fixed: true, extendLine: true }); // A and B can't be in the same place graph.pointA.onMove = function( x, y ) { return ( x != graph.pointB.coord[0] || y != graph.pointB.coord[1] ); } graph.pointB.onMove = function( x, y ) { return ( x != graph.pointA.coord[0] || y != graph.pointA.coord[1] ); } graph.pointA.toFront(); graph.pointB.toFront();

[ graph.pointA.coord, graph.pointB.coord ]

if (_.isEqual(guess, [[-5, 5], [5, 5]])) { return ""; } var slope = ( guess[1][1] - guess[0][1] ) / ( guess[1][0] - guess[0][0] ); var yint = slope * ( 0 - guess[0][0] ) + guess[0][1]; return abs( SLOPE - slope ) < 0.001 && abs( YINT - yint ) < 0.001;

graph.pointA.setCoord( guess[0] ); graph.pointB.setCoord( guess[1] ); graph.line1.transform( true );