{"id":"cb_question_bank:b10def61-855e-444f-a96c-3446455415eb","source_id":"cb_question_bank","source_item_id":"b10def61-855e-444f-a96c-3446455415eb","external_id":"8f4b667c-7537-4495-ba4d-4549d1e62f85","ibn":null,"question_id":"a35c7164","program":"SAT","module":"math","domain_code":"H","domain":"Algebra","skill_code":"H.C.","skill":"Linear equations in two variables","difficulty":"H","score_band":7,"answer_type":"mcq","answer_source":"key","layout_warning":null,"form_id":null,"module_number":null,"position":null,"stem":"5 x plus 7 y equals 1 a x plus b y equals 1 In the given pair of equations, a and b are constants. The graph of this pair of equations in the xy -plane is a pair of perpendicular lines. Which of the following pairs of equations also represents a pair of perpendicular lines?","stimulus":"","stimulus_html":null,"stem_html":"<p style=\"text-align: center;\"><math alttext=\"5 x plus 7 y equals 1\"><mrow>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mn>5</mn>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>+</mo>\n\t\t<mrow>\n\t\t\t<mn>7</mn>\n\t\t\t<mi>y</mi>\n\t\t</mrow>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>1</mn>\n</mrow>\n</math></p>\n<p style=\"text-align: center;\"><math alttext=\"a x plus b y equals 1\"><mrow>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mi>a</mi>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>+</mo>\n\t\t<mrow>\n\t\t\t<mi>b</mi>\n\t\t\t<mi>y</mi>\n\t\t</mrow>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>1</mn>\n</mrow>\n</math></p>\n<p style=\"text-align: left;\">In the given pair of equations, <math alttext=\"a\"><mi>a</mi>\n</math> and <math alttext=\"b\"><mi>b</mi>\n</math> are constants. The graph of this pair of equations in the <em>xy</em>-plane is a pair of perpendicular lines. Which of the following pairs of equations also represents a pair of perpendicular lines?</p>","rationale_html":"<p style=\"text-align: left;\">Choice B is correct. Two lines are perpendicular if their slopes are negative reciprocals, meaning that the slope of the first line is equal to <math alttext=\"negative 1\"><mo>-</mo><mn>1</mn>\n</math> divided by the slope of the second line. Each equation in the given pair of equations can be written in slope-intercept form, <math alttext=\"y equals m x plus b\"><mrow>\n\t<mi>y</mi>\n\t<mo>=</mo>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mi>m</mi>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>+</mo>\n\t\t<mi>b</mi>\n\t</mrow>\n</mrow>\n</math>, where <math alttext=\"m\"><mi>m</mi>\n</math> is the slope of the graph of the equation in the <em>xy</em>-plane and <math alttext=\"left parenthesis 0 comma b right parenthesis\"><mfenced><mrow><mn>0</mn><mo>,</mo><mi>b</mi></mrow></mfenced></math> is the <em>y</em>-intercept. For the first equation, <math alttext=\"5 x plus 7 y equals 1\"><mrow>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mn>5</mn>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>+</mo>\n\t\t<mrow>\n\t\t\t<mn>7</mn>\n\t\t\t<mi>y</mi>\n\t\t</mrow>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>1</mn>\n</mrow>\n</math>, subtracting <math alttext=\"5 x\"><mrow>\n\t<mn>5</mn>\n\t<mi>x</mi>\n</mrow>\n</math> from both sides gives <math alttext=\"7 y equals minus 5 x plus 1\"><mn>7</mn><mi>y</mi><mo>=</mo><mo>-</mo><mn>5</mn><mi>x</mi><mo>+</mo><mn>1</mn></math>, and dividing both sides of this equation by <math alttext=\"7\"><mn>7</mn>\n</math> gives <math alttext=\"y equals minus five sevenths x plus one seventh\"><mi>y</mi><mo>=</mo><mo>-</mo><mfrac><mn>5</mn><mn>7</mn></mfrac><mi>x</mi><mo>+</mo><mfrac><mn>1</mn><mn>7</mn></mfrac></math>. Therefore, the slope of the graph of this equation is <math alttext=\"negative five sevenths\"><mrow>\n\t<mo>-</mo>\n\t<mfrac>\n\t\t<mn>5</mn>\n\t\t<mn>7</mn>\n\t</mfrac>\n</mrow>\n</math>. For the second equation, <math alttext=\"a x plus b y equals 1\"><mrow>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mi>a</mi>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>+</mo>\n\t\t<mrow>\n\t\t\t<mi>b</mi>\n\t\t\t<mi>y</mi>\n\t\t</mrow>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>1</mn>\n</mrow>\n</math>, subtracting <math alttext=\"a x\"><mrow>\n\t<mi>a</mi>\n\t<mi>x</mi>\n</mrow>\n</math> from both sides gives <math alttext=\"b y equals minus a x plus 1\"><mrow>\n\t<mrow>\n\t\t<mi>b</mi>\n\t\t<mi>y</mi>\n\t</mrow>\n\t<mo>=</mo>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mo>-</mo>\n\t\t\t<mi>a</mi>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>+</mo>\n\t\t<mn>1</mn>\n\t</mrow>\n</mrow>\n</math>, and dividing both sides of this equation by <math alttext=\"b\"><mi>b</mi>\n</math> gives&nbsp;<math alttext=\"y equals minus StartFraction a Over b EndFraction x plus StartFraction 1 Over b EndFraction\"><mi>y</mi><mo>=</mo><mo>-</mo><mfrac><mi>a</mi><mi>b</mi></mfrac><mi>x</mi><mo>+</mo><mfrac><mn>1</mn><mi>b</mi></mfrac></math>. Therefore, the slope of the graph of this equation is <math alttext=\"minus StartFraction a Over b EndFraction\"><mrow>\n\t<mo>-</mo>\n\t<mfrac>\n\t\t<mi>a</mi>\n\t\t<mi>b</mi>\n\t</mfrac>\n</mrow>\n</math>. Since the graph of the given pair of equations is a pair of perpendicular lines, the slope of the graph of the second equation, <math alttext=\"minus StartFraction a Over b EndFraction\"><mrow>\n\t<mo>-</mo>\n\t<mfrac>\n\t\t<mi>a</mi>\n\t\t<mi>b</mi>\n\t</mfrac>\n</mrow>\n</math>, must be the negative reciprocal of the slope of the graph of the first equation, <math alttext=\"negative five sevenths\"><mrow>\n\t<mo>-</mo>\n\t<mfrac>\n\t\t<mn>5</mn>\n\t\t<mn>7</mn>\n\t</mfrac>\n</mrow>\n</math>. The negative reciprocal of <math alttext=\"negative five sevenths\"><mrow>\n\t<mo>-</mo>\n\t<mfrac>\n\t\t<mn>5</mn>\n\t\t<mn>7</mn>\n\t</mfrac>\n</mrow>\n</math> is&nbsp; <math alttext=\"StartFraction negative 1 Over left parenthesis negative five sevenths right parenthesis EndFraction\"><mfrac><mrow><mo>-</mo><mn>1</mn></mrow><mstyle displaystyle=\"true\"><mfenced><mrow><mo>-</mo><mfrac><mn>5</mn><mn>7</mn></mfrac></mrow></mfenced></mstyle></mfrac></math>, or <math alttext=\"seven fifths\"><mfrac>\n\t<mn>7</mn>\n\t<mn>5</mn>\n</mfrac>\n</math>. Therefore, <math alttext=\"minus StartFraction a Over b EndFraction equals seven fifths\"><mrow>\n\t<mrow>\n\t\t<mo>-</mo>\n\t\t<mfrac>\n\t\t\t<mi>a</mi>\n\t\t\t<mi>b</mi>\n\t\t</mfrac>\n\t</mrow>\n\t<mo>=</mo>\n\t<mfrac>\n\t\t<mn>7</mn>\n\t\t<mn>5</mn>\n\t</mfrac>\n</mrow>\n</math>,&nbsp;or <math alttext=\"StartFraction a Over b EndFraction equals negative seven fifths\"><mrow>\n\t<mfrac>\n\t\t\t<mi>a</mi>\n\t\t\t<mi>b</mi>\n\t\t</mfrac>\n\t<mo>=</mo>\n\t<mrow>\n\t\t<mo>-</mo>\n\t\t<mfrac>\n\t\t\t<mn>7</mn>\n\t\t\t<mn>5</mn>\n\t\t</mfrac>\n\t</mrow>\n</mrow>\n</math>. Similarly, rewriting the equations in choice B in slope-intercept form yields&nbsp;<math alttext=\"y equals minus StartFraction 10 Over 7 EndFraction x plus one seventh\"><mi>y</mi><mo>=</mo><mo>-</mo><mfrac><mn>10</mn><mn>7</mn></mfrac><mi>x</mi><mo>+</mo><mfrac><mn>1</mn><mn>7</mn></mfrac></math> and&nbsp;<math alttext=\"y equals minus StartFraction a Over 2 b EndFraction x plus StartFraction 1 Over 2 b EndFraction\"><mi>y</mi><mo>=</mo><mo>-</mo><mfrac><mi>a</mi><mrow><mn>2</mn><mi>b</mi></mrow></mfrac><mi>x</mi><mo>+</mo><mfrac><mn>1</mn><mrow><mn>2</mn><mi>b</mi></mrow></mfrac></math>. It follows that the slope of the graph of the first equation in choice B is <math alttext=\"negative StartFraction 10 Over 7 EndFraction\"><mrow>\n\t<mo>-</mo>\n\t<mfrac>\n\t\t<mn>10</mn>\n\t\t<mn>7</mn>\n\t</mfrac>\n</mrow>\n</math> and the slope of the graph of the second equation in choice B is <math alttext=\"minus StartFraction a Over 2 b EndFraction\"><mrow>\n\t<mo>-</mo>\n\t<mfrac>\n\t\t<mi>a</mi>\n\t\t<mrow>\n\t\t\t<mn>2</mn>\n\t\t\t<mi>b</mi>\n\t\t</mrow>\n\t</mfrac>\n</mrow>\n</math>. Since <math alttext=\"StartFraction a Over b EndFraction equals negative seven fifths\"><mrow>\n\t<mfrac>\n\t\t\t<mi>a</mi>\n\t\t\t<mi>b</mi>\n\t\t</mfrac>\n\t<mo>=</mo>\n\t<mrow>\n\t\t<mo>-</mo>\n\t\t<mfrac>\n\t\t\t<mn>7</mn>\n\t\t\t<mn>5</mn>\n\t\t</mfrac>\n\t</mrow>\n</mrow>\n</math>, <math alttext=\"minus StartFraction a Over 2 b EndFraction\"><mrow>\n\t<mo>-</mo>\n\t<mfrac>\n\t\t<mi>a</mi>\n\t\t<mrow>\n\t\t\t<mn>2</mn>\n\t\t\t<mi>b</mi>\n\t\t</mrow>\n\t</mfrac>\n</mrow>\n</math> is equal to <math alttext=\"minus left parenthesis one half right parenthesis left parenthesis negative seven fifths right parenthesis\"><mo>-</mo><mfenced><mfrac><mn>1</mn><mn>2</mn></mfrac></mfenced><mfenced><mrow><mo>-</mo><mfrac><mn>7</mn><mn>5</mn></mfrac></mrow></mfenced></math>, or <math alttext=\"seven tenths\"><mfrac>\n\t<mn>7</mn>\n\t<mn>10</mn>\n</mfrac>\n</math>. Since <math alttext=\"seven tenths\"><mfrac>\n\t<mn>7</mn>\n\t<mn>10</mn>\n</mfrac>\n</math> is the negative reciprocal of <math alttext=\"negative StartFraction 10 Over 7 EndFraction\"><mrow>\n\t<mo>-</mo>\n\t<mfrac>\n\t\t<mn>10</mn>\n\t\t<mn>7</mn>\n\t</mfrac>\n</mrow>\n</math>, the pair of equations in choice B represents a pair of perpendicular lines.</p>\n<p style=\"text-align: left;\">Choice A is incorrect and may result from conceptual or calculation errors.</p>\n<p style=\"text-align: left;\">Choice C is incorrect and may result from conceptual or calculation errors.</p>\n<p style=\"text-align: left;\">Choice D is incorrect and may result from conceptual or calculation errors.</p>","correct_answer":["B"],"has_media":false,"has_mathml":true,"has_table":false,"source_created_at":"2023-08-02T20:25:59.816000+00:00","source_updated_at":"2023-08-02T20:25:59.816000+00:00","options":[{"label":"A","content_html":"<p style=\"text-align: left;\"><math alttext=\"10 x plus 7 y equals 1\"><mrow>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mn>10</mn>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>+</mo>\n\t\t<mrow>\n\t\t\t<mn>7</mn>\n\t\t\t<mi>y</mi>\n\t\t</mrow>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>1</mn>\n</mrow>\n</math></p>\n<p style=\"text-align: left;\"><math alttext=\"a x minus 2 b y equals 1\"><mrow>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mi>a</mi>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>-</mo>\n\t\t<mrow>\n\t\t\t<mn>2</mn>\n\t\t\t<mi>b</mi>\n\t\t\t<mi>y</mi>\n\t\t</mrow>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>1</mn>\n</mrow>\n</math></p>","ord":0},{"label":"B","content_html":"<p style=\"text-align: left;\"><math alttext=\"10 x plus 7 y equals 1\"><mrow>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mn>10</mn>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>+</mo>\n\t\t<mrow>\n\t\t\t<mn>7</mn>\n\t\t\t<mi>y</mi>\n\t\t</mrow>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>1</mn>\n</mrow>\n</math></p>\n<p style=\"text-align: left;\"><math alttext=\"a x plus 2 b y equals 1\"><mrow>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mi>a</mi>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>+</mo>\n\t\t<mrow>\n\t\t\t<mn>2</mn>\n\t\t\t<mi>b</mi>\n\t\t\t<mi>y</mi>\n\t\t</mrow>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>1</mn>\n</mrow>\n</math></p>","ord":1},{"label":"C","content_html":"<p style=\"text-align: left;\"><math alttext=\"10 x plus 7 y equals 1\"><mrow>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mn>10</mn>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>+</mo>\n\t\t<mrow>\n\t\t\t<mn>7</mn>\n\t\t\t<mi>y</mi>\n\t\t</mrow>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>1</mn>\n</mrow>\n</math></p>\n<p style=\"text-align: left;\"><math alttext=\"2 a x plus b y equals 1\"><mrow>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mn>2</mn>\n\t\t\t<mi>a</mi>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>+</mo>\n\t\t<mrow>\n\t\t\t<mi>b</mi>\n\t\t\t<mi>y</mi>\n\t\t</mrow>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>1</mn>\n</mrow>\n</math></p>","ord":2},{"label":"D","content_html":"<p style=\"text-align: left;\"><math alttext=\"5 x minus 7 y equals 1\"><mrow>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mn>5</mn>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>-</mo>\n\t\t<mrow>\n\t\t\t<mn>7</mn>\n\t\t\t<mi>y</mi>\n\t\t</mrow>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>1</mn>\n</mrow>\n</math></p>\n<p style=\"text-align: left;\"><math alttext=\"a x plus b y equals 1\"><mrow>\n\t<mrow>\n\t\t<mrow>\n\t\t\t<mi>a</mi>\n\t\t\t<mi>x</mi>\n\t\t</mrow>\n\t\t<mo>+</mo>\n\t\t<mrow>\n\t\t\t<mi>b</mi>\n\t\t\t<mi>y</mi>\n\t\t</mrow>\n\t</mrow>\n\t<mo>=</mo>\n\t<mn>1</mn>\n</mrow>\n</math></p>","ord":3}],"attribution":{"source_id":"cb_question_bank","source_name":"College Board digital SAT question bank (community dump)","rights_holder":"College Board","canonical_url":"https://satsuitequestionbank.collegeboard.org/","retrieved_from":"https://raw.githubusercontent.com/mdn522/sat-question-bank/main/data/cb-digital-questions.json","attribution":"SAT practice questions © College Board, from the official SAT Suite Question Bank. Retrieved via the community dump at github.com/mdn522/sat-question-bank.","disclaimer":"SAT® and PSAT/NMSQT® are trademarks registered by the College Board, which is not affiliated with, does not sponsor, and does not endorse this project.","license":"College Board copyright; source repository carries no licence. Local use only.","redistributable":false,"item_count":2017}}